how to find the equation of a circle given points of the circleFind the equation of a circle given two points...
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how to find the equation of a circle given points of the circle
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how to find the equation of a circle given points of the circle
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$begingroup$
Find the equation of the circle which passes through points $P(-6,5)$, $Q(2,1)$ and has its centre lying on the line $y=x+4$
I would appreciate any working given, as I do not understand how to find the equation of the circle from this information alone.
geometry circles
edited 2 hours ago
dantopa
6,74442345
asked 3 hours ago
injustice fellowinjustice fellow
312
$endgroup$
add a comment |
$begingroup$
Find the equation of the circle which passes through points $P(-6,5)$, $Q(2,1)$ and has its centre lying on the line $y=x+4$
I would appreciate any working given, as I do not understand how to find the equation of the circle from this information alone.
geometry circles
edited 2 hours ago
dantopa
6,74442345
asked 3 hours ago
injustice fellowinjustice fellow
312
$endgroup$
$begingroup$
Use the equation of a circle with radius $r$ (unknown) and center $(x_0, y_0)$ (unknown).
$endgroup$
– Mauro ALLEGRANZA
3 hours ago
$begingroup$
Then solve te system of equations obtained from the previous one (3 unknown) knowing that points P and Q satisfy the equation of the circle and that the coordinates of the center satisfy the equation of the line.
$endgroup$
– Mauro ALLEGRANZA
3 hours ago
add a comment |
$begingroup$
Find the equation of the circle which passes through points $P(-6,5)$, $Q(2,1)$ and has its centre lying on the line $y=x+4$
I would appreciate any working given, as I do not understand how to find the equation of the circle from this information alone.
geometry circles
edited 2 hours ago
dantopa
6,74442345
asked 3 hours ago
injustice fellowinjustice fellow
312
$endgroup$
Find the equation of the circle which passes through points $P(-6,5)$, $Q(2,1)$ and has its centre lying on the line $y=x+4$
I would appreciate any working given, as I do not understand how to find the equation of the circle from this information alone.
geometry circles
geometry circles
edited 2 hours ago
dantopa
6,74442345
asked 3 hours ago
injustice fellowinjustice fellow
312
edited 2 hours ago
dantopa
6,74442345
asked 3 hours ago
injustice fellowinjustice fellow
312
edited 2 hours ago
dantopa
6,74442345
edited 2 hours ago
dantopa
6,74442345
edited 2 hours ago
dantopa
6,74442345
6,74442345
asked 3 hours ago
injustice fellowinjustice fellow
312
asked 3 hours ago
injustice fellowinjustice fellow
312
asked 3 hours ago
injustice fellowinjustice fellow
312
312
$begingroup$
Use the equation of a circle with radius $r$ (unknown) and center $(x_0, y_0)$ (unknown).
$endgroup$
– Mauro ALLEGRANZA
3 hours ago
$begingroup$
Then solve te system of equations obtained from the previous one (3 unknown) knowing that points P and Q satisfy the equation of the circle and that the coordinates of the center satisfy the equation of the line.
$endgroup$
– Mauro ALLEGRANZA
3 hours ago
add a comment |
$begingroup$
Use the equation of a circle with radius $r$ (unknown) and center $(x_0, y_0)$ (unknown).
$endgroup$
– Mauro ALLEGRANZA
3 hours ago
$begingroup$
Then solve te system of equations obtained from the previous one (3 unknown) knowing that points P and Q satisfy the equation of the circle and that the coordinates of the center satisfy the equation of the line.
$endgroup$
– Mauro ALLEGRANZA
3 hours ago
$begingroup$
Use the equation of a circle with radius $r$ (unknown) and center $(x_0, y_0)$ (unknown).
$endgroup$
– Mauro ALLEGRANZA
3 hours ago
$begingroup$
Use the equation of a circle with radius $r$ (unknown) and center $(x_0, y_0)$ (unknown).
$endgroup$
– Mauro ALLEGRANZA
3 hours ago
$begingroup$
Then solve te system of equations obtained from the previous one (3 unknown) knowing that points P and Q satisfy the equation of the circle and that the coordinates of the center satisfy the equation of the line.
$endgroup$
– Mauro ALLEGRANZA
3 hours ago
$begingroup$
Then solve te system of equations obtained from the previous one (3 unknown) knowing that points P and Q satisfy the equation of the circle and that the coordinates of the center satisfy the equation of the line.
$endgroup$
– Mauro ALLEGRANZA
3 hours ago
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Hint:
Since the centre of the circle will also lie on the perpendicular bisector of the given two points, so you can find the coordinates as the intersection of the bisector and $y=x+4$.
The radius would be the distance from the centre to any of the given points.
P.S. In this case, the perpendicular bisector is different from the given line. If it weren't so, then the given points would be the ends of a diameter, from which you could easily find the center lying midway.(Thanks to @TonyK for pointing out the inaccuracy)
P.S. If the perpendicular bisector is the given line, then there won't be a unique circle satisfying the conditions, but you'll get a family of circles.
Image for the P.S. part, showing two of the possible circles if the given line was also a perpendicular bisector of the given points
answered 2 hours ago
EagleEagle
19313
$endgroup$
2
$begingroup$
+1 for being the only answer so far to use the geometry of the thing. But your PS is wrong: if the perpendicular bisector is coincident with the given line, then the centre of the circle can be any point on the line.
$endgroup$
– TonyK
2 hours ago
$begingroup$
@TonyK yeah thanks for pointing it out. I'll edit it.
$endgroup$
– Eagle
2 hours ago
$begingroup$
@TonyK And how do we get the equation $$(x-a)^2+(y-b)^2 = r^2$$ without geometry?
$endgroup$
– Maria Mazur
2 hours ago
$begingroup$
@MariaMazur: What I really meant was that you can solve this problem simply by finding the intersection of two straight lines. There is no need for quadratic equations.
$endgroup$
– TonyK
2 hours ago
add a comment |
$begingroup$
The equation of circle is $$(x-a)^2+(y-b)^2 = r^2$$where $(a,b)$ is a center of a circle and $r$ radius of a circle. Since ceneter is on a line $y=x+4$ we have $b=a+4$ so we have now:
$$(x-a)^2+(y-a-4)^2 = r^2$$
Now put both points in to this equation and you get:
$$ P: ;;;(-6-a)^2+(1-a)^2 = r^2$$
and
$$ Q: ;;;(2-a)^2+(-a-3)^2 = r^2$$
Now solve this system...
answered 2 hours ago
Maria MazurMaria Mazur
50.7k1362126
$endgroup$
add a comment |
$begingroup$
Let the center be at $(t,t+4)$. We express that the circle is through the given points:
$$r^2=(t+6)^2+(t+4-5)^2=(t-2)^2+(t+4-1)^2.$$
After simplification,
$$8t+24=0$$ and you are nearly done.
$t=-3$, center $(-3,1)$, radius $5$.
answered 2 hours ago
Yves DaoustYves Daoust
134k676232
$endgroup$
1
$begingroup$
(+).................
$endgroup$
– E.H.E
2 hours ago
add a comment |
$begingroup$
Let the center be $C(x_0,y_0)$
As P and Q lie on the circle, CP = CQ = radius of the circle.
$$CP = CQ$$
$$CP^2 = CQ^2$$
$$(x_1-(-6))^2 + (y_1 -5)^2 = (x_1-2)^2 + (y_1 -1)^2$$
$$x_1^2 + 12x_1 +36 + y_1^2 - 10y_1 +25 = x_1^2 -4x_1 +4 + y_1^2 - 2y_1 +1$$
$$12x_1+36 - 10y_1 +25 = -4x_1 + 4 -2y_1 + 1$$
$$16x_1 - 8y_1 +56 = 0$$ or $$ 2x_1 - y_1 + 7 =0$$
We also have$$y_1 = x_1+4$$
So,$$2x_1 - x_1 - 4+ 7 = 0 $$
$$x_1 = -3$$
$$y_1 = x_1+4 = 1$$
Centre $C = C(-3,1)$
Radius $r = CQ = sqrt{(-3-2)^2 + (1-1)^2} = 5$$
Thus the equation is,$$(x-x_1)^2 + (y-y_1)^2 = r^2$$
or$$(x+3)^2+(y-1)^2 = 25$$
answered 2 hours ago
Ak19Ak19
1425
New contributor
Ak19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Consider the circunference $C$
$$
(x + 3)^2 + (y - 1)^2 = 25
$$
We have that $P,Qin C$ (just replacing coordinates). And the center is $(-3,1)$ which is in the line $y=x+4$. Since $P,Qin C$ the center must be in the perpendicular line ($-2x+y=7$) to the one that joins $P,Q$. Since the intersection between the two lines that we have consider is a point, we deduce that the solution is unique and is the one we have proposed.
answered 2 hours ago
davidivadfuldavidivadful
14410
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Since the centre of the circle will also lie on the perpendicular bisector of the given two points, so you can find the coordinates as the intersection of the bisector and $y=x+4$.
The radius would be the distance from the centre to any of the given points.
P.S. In this case, the perpendicular bisector is different from the given line. If it weren't so, then the given points would be the ends of a diameter, from which you could easily find the center lying midway.(Thanks to @TonyK for pointing out the inaccuracy)
P.S. If the perpendicular bisector is the given line, then there won't be a unique circle satisfying the conditions, but you'll get a family of circles.
Image for the P.S. part, showing two of the possible circles if the given line was also a perpendicular bisector of the given points
answered 2 hours ago
EagleEagle
19313
$endgroup$
2
$begingroup$
+1 for being the only answer so far to use the geometry of the thing. But your PS is wrong: if the perpendicular bisector is coincident with the given line, then the centre of the circle can be any point on the line.
$endgroup$
– TonyK
2 hours ago
$begingroup$
@TonyK yeah thanks for pointing it out. I'll edit it.
$endgroup$
– Eagle
2 hours ago
$begingroup$
@TonyK And how do we get the equation $$(x-a)^2+(y-b)^2 = r^2$$ without geometry?
$endgroup$
– Maria Mazur
2 hours ago
$begingroup$
@MariaMazur: What I really meant was that you can solve this problem simply by finding the intersection of two straight lines. There is no need for quadratic equations.
$endgroup$
– TonyK
2 hours ago
add a comment |
$begingroup$
Hint:
Since the centre of the circle will also lie on the perpendicular bisector of the given two points, so you can find the coordinates as the intersection of the bisector and $y=x+4$.
The radius would be the distance from the centre to any of the given points.
P.S. In this case, the perpendicular bisector is different from the given line. If it weren't so, then the given points would be the ends of a diameter, from which you could easily find the center lying midway.(Thanks to @TonyK for pointing out the inaccuracy)
P.S. If the perpendicular bisector is the given line, then there won't be a unique circle satisfying the conditions, but you'll get a family of circles.
Image for the P.S. part, showing two of the possible circles if the given line was also a perpendicular bisector of the given points
answered 2 hours ago
EagleEagle
19313
$endgroup$
2
$begingroup$
+1 for being the only answer so far to use the geometry of the thing. But your PS is wrong: if the perpendicular bisector is coincident with the given line, then the centre of the circle can be any point on the line.
$endgroup$
– TonyK
2 hours ago
$begingroup$
@TonyK yeah thanks for pointing it out. I'll edit it.
$endgroup$
– Eagle
2 hours ago
$begingroup$
@TonyK And how do we get the equation $$(x-a)^2+(y-b)^2 = r^2$$ without geometry?
$endgroup$
– Maria Mazur
2 hours ago
$begingroup$
@MariaMazur: What I really meant was that you can solve this problem simply by finding the intersection of two straight lines. There is no need for quadratic equations.
$endgroup$
– TonyK
2 hours ago
add a comment |
$begingroup$
Hint:
Since the centre of the circle will also lie on the perpendicular bisector of the given two points, so you can find the coordinates as the intersection of the bisector and $y=x+4$.
The radius would be the distance from the centre to any of the given points.
P.S. In this case, the perpendicular bisector is different from the given line. If it weren't so, then the given points would be the ends of a diameter, from which you could easily find the center lying midway.(Thanks to @TonyK for pointing out the inaccuracy)
P.S. If the perpendicular bisector is the given line, then there won't be a unique circle satisfying the conditions, but you'll get a family of circles.
Image for the P.S. part, showing two of the possible circles if the given line was also a perpendicular bisector of the given points
answered 2 hours ago
EagleEagle
19313
$endgroup$
Hint:
Since the centre of the circle will also lie on the perpendicular bisector of the given two points, so you can find the coordinates as the intersection of the bisector and $y=x+4$.
The radius would be the distance from the centre to any of the given points.
P.S. In this case, the perpendicular bisector is different from the given line. If it weren't so, then the given points would be the ends of a diameter, from which you could easily find the center lying midway.(Thanks to @TonyK for pointing out the inaccuracy)
P.S. If the perpendicular bisector is the given line, then there won't be a unique circle satisfying the conditions, but you'll get a family of circles.
Image for the P.S. part, showing two of the possible circles if the given line was also a perpendicular bisector of the given points
answered 2 hours ago
EagleEagle
19313
edited 2 hours ago
answered 2 hours ago
EagleEagle
19313
answered 2 hours ago
EagleEagle
19313
answered 2 hours ago
EagleEagle
19313
19313
2
$begingroup$
+1 for being the only answer so far to use the geometry of the thing. But your PS is wrong: if the perpendicular bisector is coincident with the given line, then the centre of the circle can be any point on the line.
$endgroup$
– TonyK
2 hours ago
$begingroup$
@TonyK yeah thanks for pointing it out. I'll edit it.
$endgroup$
– Eagle
2 hours ago
$begingroup$
@TonyK And how do we get the equation $$(x-a)^2+(y-b)^2 = r^2$$ without geometry?
$endgroup$
– Maria Mazur
2 hours ago
$begingroup$
@MariaMazur: What I really meant was that you can solve this problem simply by finding the intersection of two straight lines. There is no need for quadratic equations.
$endgroup$
– TonyK
2 hours ago
add a comment |
2
$begingroup$
+1 for being the only answer so far to use the geometry of the thing. But your PS is wrong: if the perpendicular bisector is coincident with the given line, then the centre of the circle can be any point on the line.
$endgroup$
– TonyK
2 hours ago
$begingroup$
@TonyK yeah thanks for pointing it out. I'll edit it.
$endgroup$
– Eagle
2 hours ago
$begingroup$
@TonyK And how do we get the equation $$(x-a)^2+(y-b)^2 = r^2$$ without geometry?
$endgroup$
– Maria Mazur
2 hours ago
$begingroup$
@MariaMazur: What I really meant was that you can solve this problem simply by finding the intersection of two straight lines. There is no need for quadratic equations.
$endgroup$
– TonyK
2 hours ago
2
2
$begingroup$
+1 for being the only answer so far to use the geometry of the thing. But your PS is wrong: if the perpendicular bisector is coincident with the given line, then the centre of the circle can be any point on the line.
$endgroup$
– TonyK
2 hours ago
$begingroup$
+1 for being the only answer so far to use the geometry of the thing. But your PS is wrong: if the perpendicular bisector is coincident with the given line, then the centre of the circle can be any point on the line.
$endgroup$
– TonyK
2 hours ago
$begingroup$
@TonyK yeah thanks for pointing it out. I'll edit it.
$endgroup$
– Eagle
2 hours ago
$begingroup$
@TonyK yeah thanks for pointing it out. I'll edit it.
$endgroup$
– Eagle
2 hours ago
$begingroup$
@TonyK And how do we get the equation $$(x-a)^2+(y-b)^2 = r^2$$ without geometry?
$endgroup$
– Maria Mazur
2 hours ago
$begingroup$
@TonyK And how do we get the equation $$(x-a)^2+(y-b)^2 = r^2$$ without geometry?
$endgroup$
– Maria Mazur
2 hours ago
$begingroup$
@MariaMazur: What I really meant was that you can solve this problem simply by finding the intersection of two straight lines. There is no need for quadratic equations.
$endgroup$
– TonyK
2 hours ago
$begingroup$
@MariaMazur: What I really meant was that you can solve this problem simply by finding the intersection of two straight lines. There is no need for quadratic equations.
$endgroup$
– TonyK
2 hours ago
add a comment |
$begingroup$
The equation of circle is $$(x-a)^2+(y-b)^2 = r^2$$where $(a,b)$ is a center of a circle and $r$ radius of a circle. Since ceneter is on a line $y=x+4$ we have $b=a+4$ so we have now:
$$(x-a)^2+(y-a-4)^2 = r^2$$
Now put both points in to this equation and you get:
$$ P: ;;;(-6-a)^2+(1-a)^2 = r^2$$
and
$$ Q: ;;;(2-a)^2+(-a-3)^2 = r^2$$
Now solve this system...
answered 2 hours ago
Maria MazurMaria Mazur
50.7k1362126
$endgroup$
add a comment |
$begingroup$
The equation of circle is $$(x-a)^2+(y-b)^2 = r^2$$where $(a,b)$ is a center of a circle and $r$ radius of a circle. Since ceneter is on a line $y=x+4$ we have $b=a+4$ so we have now:
$$(x-a)^2+(y-a-4)^2 = r^2$$
Now put both points in to this equation and you get:
$$ P: ;;;(-6-a)^2+(1-a)^2 = r^2$$
and
$$ Q: ;;;(2-a)^2+(-a-3)^2 = r^2$$
Now solve this system...
answered 2 hours ago
Maria MazurMaria Mazur
50.7k1362126
$endgroup$
add a comment |
$begingroup$
The equation of circle is $$(x-a)^2+(y-b)^2 = r^2$$where $(a,b)$ is a center of a circle and $r$ radius of a circle. Since ceneter is on a line $y=x+4$ we have $b=a+4$ so we have now:
$$(x-a)^2+(y-a-4)^2 = r^2$$
Now put both points in to this equation and you get:
$$ P: ;;;(-6-a)^2+(1-a)^2 = r^2$$
and
$$ Q: ;;;(2-a)^2+(-a-3)^2 = r^2$$
Now solve this system...
answered 2 hours ago
Maria MazurMaria Mazur
50.7k1362126
$endgroup$
The equation of circle is $$(x-a)^2+(y-b)^2 = r^2$$where $(a,b)$ is a center of a circle and $r$ radius of a circle. Since ceneter is on a line $y=x+4$ we have $b=a+4$ so we have now:
$$(x-a)^2+(y-a-4)^2 = r^2$$
Now put both points in to this equation and you get:
$$ P: ;;;(-6-a)^2+(1-a)^2 = r^2$$
and
$$ Q: ;;;(2-a)^2+(-a-3)^2 = r^2$$
Now solve this system...
answered 2 hours ago
Maria MazurMaria Mazur
50.7k1362126
answered 2 hours ago
Maria MazurMaria Mazur
50.7k1362126
answered 2 hours ago
Maria MazurMaria Mazur
50.7k1362126
answered 2 hours ago
Maria MazurMaria Mazur
50.7k1362126
50.7k1362126
add a comment |
add a comment |
$begingroup$
Let the center be at $(t,t+4)$. We express that the circle is through the given points:
$$r^2=(t+6)^2+(t+4-5)^2=(t-2)^2+(t+4-1)^2.$$
After simplification,
$$8t+24=0$$ and you are nearly done.
$t=-3$, center $(-3,1)$, radius $5$.
answered 2 hours ago
Yves DaoustYves Daoust
134k676232
$endgroup$
1
$begingroup$
(+).................
$endgroup$
– E.H.E
2 hours ago
add a comment |
$begingroup$
Let the center be at $(t,t+4)$. We express that the circle is through the given points:
$$r^2=(t+6)^2+(t+4-5)^2=(t-2)^2+(t+4-1)^2.$$
After simplification,
$$8t+24=0$$ and you are nearly done.
$t=-3$, center $(-3,1)$, radius $5$.
answered 2 hours ago
Yves DaoustYves Daoust
134k676232
$endgroup$
1
$begingroup$
(+).................
$endgroup$
– E.H.E
2 hours ago
add a comment |
$begingroup$
Let the center be at $(t,t+4)$. We express that the circle is through the given points:
$$r^2=(t+6)^2+(t+4-5)^2=(t-2)^2+(t+4-1)^2.$$
After simplification,
$$8t+24=0$$ and you are nearly done.
$t=-3$, center $(-3,1)$, radius $5$.
answered 2 hours ago
Yves DaoustYves Daoust
134k676232
$endgroup$
Let the center be at $(t,t+4)$. We express that the circle is through the given points:
$$r^2=(t+6)^2+(t+4-5)^2=(t-2)^2+(t+4-1)^2.$$
After simplification,
$$8t+24=0$$ and you are nearly done.
$t=-3$, center $(-3,1)$, radius $5$.
answered 2 hours ago
Yves DaoustYves Daoust
134k676232
edited 1 hour ago
answered 2 hours ago
Yves DaoustYves Daoust
134k676232
answered 2 hours ago
Yves DaoustYves Daoust
134k676232
answered 2 hours ago
Yves DaoustYves Daoust
134k676232
134k676232
1
$begingroup$
(+).................
$endgroup$
– E.H.E
2 hours ago
add a comment |
1
$begingroup$
(+).................
$endgroup$
– E.H.E
2 hours ago
1
1
$begingroup$
(+).................
$endgroup$
– E.H.E
2 hours ago
$begingroup$
(+).................
$endgroup$
– E.H.E
2 hours ago
add a comment |
$begingroup$
Let the center be $C(x_0,y_0)$
As P and Q lie on the circle, CP = CQ = radius of the circle.
$$CP = CQ$$
$$CP^2 = CQ^2$$
$$(x_1-(-6))^2 + (y_1 -5)^2 = (x_1-2)^2 + (y_1 -1)^2$$
$$x_1^2 + 12x_1 +36 + y_1^2 - 10y_1 +25 = x_1^2 -4x_1 +4 + y_1^2 - 2y_1 +1$$
$$12x_1+36 - 10y_1 +25 = -4x_1 + 4 -2y_1 + 1$$
$$16x_1 - 8y_1 +56 = 0$$ or $$ 2x_1 - y_1 + 7 =0$$
We also have$$y_1 = x_1+4$$
So,$$2x_1 - x_1 - 4+ 7 = 0 $$
$$x_1 = -3$$
$$y_1 = x_1+4 = 1$$
Centre $C = C(-3,1)$
Radius $r = CQ = sqrt{(-3-2)^2 + (1-1)^2} = 5$$
Thus the equation is,$$(x-x_1)^2 + (y-y_1)^2 = r^2$$
or$$(x+3)^2+(y-1)^2 = 25$$
answered 2 hours ago
Ak19Ak19
1425
New contributor
Ak19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
Let the center be $C(x_0,y_0)$
As P and Q lie on the circle, CP = CQ = radius of the circle.
$$CP = CQ$$
$$CP^2 = CQ^2$$
$$(x_1-(-6))^2 + (y_1 -5)^2 = (x_1-2)^2 + (y_1 -1)^2$$
$$x_1^2 + 12x_1 +36 + y_1^2 - 10y_1 +25 = x_1^2 -4x_1 +4 + y_1^2 - 2y_1 +1$$
$$12x_1+36 - 10y_1 +25 = -4x_1 + 4 -2y_1 + 1$$
$$16x_1 - 8y_1 +56 = 0$$ or $$ 2x_1 - y_1 + 7 =0$$
We also have$$y_1 = x_1+4$$
So,$$2x_1 - x_1 - 4+ 7 = 0 $$
$$x_1 = -3$$
$$y_1 = x_1+4 = 1$$
Centre $C = C(-3,1)$
Radius $r = CQ = sqrt{(-3-2)^2 + (1-1)^2} = 5$$
Thus the equation is,$$(x-x_1)^2 + (y-y_1)^2 = r^2$$
or$$(x+3)^2+(y-1)^2 = 25$$
answered 2 hours ago
Ak19Ak19
1425
New contributor
Ak19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let the center be $C(x_0,y_0)$
As P and Q lie on the circle, CP = CQ = radius of the circle.
$$CP = CQ$$
$$CP^2 = CQ^2$$
$$(x_1-(-6))^2 + (y_1 -5)^2 = (x_1-2)^2 + (y_1 -1)^2$$
$$x_1^2 + 12x_1 +36 + y_1^2 - 10y_1 +25 = x_1^2 -4x_1 +4 + y_1^2 - 2y_1 +1$$
$$12x_1+36 - 10y_1 +25 = -4x_1 + 4 -2y_1 + 1$$
$$16x_1 - 8y_1 +56 = 0$$ or $$ 2x_1 - y_1 + 7 =0$$
We also have$$y_1 = x_1+4$$
So,$$2x_1 - x_1 - 4+ 7 = 0 $$
$$x_1 = -3$$
$$y_1 = x_1+4 = 1$$
Centre $C = C(-3,1)$
Radius $r = CQ = sqrt{(-3-2)^2 + (1-1)^2} = 5$$
Thus the equation is,$$(x-x_1)^2 + (y-y_1)^2 = r^2$$
or$$(x+3)^2+(y-1)^2 = 25$$
answered 2 hours ago
Ak19Ak19
1425
New contributor
Ak19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let the center be $C(x_0,y_0)$
As P and Q lie on the circle, CP = CQ = radius of the circle.
$$CP = CQ$$
$$CP^2 = CQ^2$$
$$(x_1-(-6))^2 + (y_1 -5)^2 = (x_1-2)^2 + (y_1 -1)^2$$
$$x_1^2 + 12x_1 +36 + y_1^2 - 10y_1 +25 = x_1^2 -4x_1 +4 + y_1^2 - 2y_1 +1$$
$$12x_1+36 - 10y_1 +25 = -4x_1 + 4 -2y_1 + 1$$
$$16x_1 - 8y_1 +56 = 0$$ or $$ 2x_1 - y_1 + 7 =0$$
We also have$$y_1 = x_1+4$$
So,$$2x_1 - x_1 - 4+ 7 = 0 $$
$$x_1 = -3$$
$$y_1 = x_1+4 = 1$$
Centre $C = C(-3,1)$
Radius $r = CQ = sqrt{(-3-2)^2 + (1-1)^2} = 5$$
Thus the equation is,$$(x-x_1)^2 + (y-y_1)^2 = r^2$$
or$$(x+3)^2+(y-1)^2 = 25$$
answered 2 hours ago
Ak19Ak19
1425
New contributor
Ak19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Ak19Ak19
1425
New contributor
Ak19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Ak19Ak19
1425
answered 2 hours ago
Ak19Ak19
1425
1425
New contributor
Ak19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ak19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ak19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
Consider the circunference $C$
$$
(x + 3)^2 + (y - 1)^2 = 25
$$
We have that $P,Qin C$ (just replacing coordinates). And the center is $(-3,1)$ which is in the line $y=x+4$. Since $P,Qin C$ the center must be in the perpendicular line ($-2x+y=7$) to the one that joins $P,Q$. Since the intersection between the two lines that we have consider is a point, we deduce that the solution is unique and is the one we have proposed.
answered 2 hours ago
davidivadfuldavidivadful
14410
$endgroup$
add a comment |
$begingroup$
Consider the circunference $C$
$$
(x + 3)^2 + (y - 1)^2 = 25
$$
We have that $P,Qin C$ (just replacing coordinates). And the center is $(-3,1)$ which is in the line $y=x+4$. Since $P,Qin C$ the center must be in the perpendicular line ($-2x+y=7$) to the one that joins $P,Q$. Since the intersection between the two lines that we have consider is a point, we deduce that the solution is unique and is the one we have proposed.
answered 2 hours ago
davidivadfuldavidivadful
14410
$endgroup$
add a comment |
$begingroup$
Consider the circunference $C$
$$
(x + 3)^2 + (y - 1)^2 = 25
$$
We have that $P,Qin C$ (just replacing coordinates). And the center is $(-3,1)$ which is in the line $y=x+4$. Since $P,Qin C$ the center must be in the perpendicular line ($-2x+y=7$) to the one that joins $P,Q$. Since the intersection between the two lines that we have consider is a point, we deduce that the solution is unique and is the one we have proposed.
answered 2 hours ago
davidivadfuldavidivadful
14410
$endgroup$
Consider the circunference $C$
$$
(x + 3)^2 + (y - 1)^2 = 25
$$
We have that $P,Qin C$ (just replacing coordinates). And the center is $(-3,1)$ which is in the line $y=x+4$. Since $P,Qin C$ the center must be in the perpendicular line ($-2x+y=7$) to the one that joins $P,Q$. Since the intersection between the two lines that we have consider is a point, we deduce that the solution is unique and is the one we have proposed.
answered 2 hours ago
davidivadfuldavidivadful
14410
answered 2 hours ago
davidivadfuldavidivadful
14410
answered 2 hours ago
davidivadfuldavidivadful
14410
answered 2 hours ago
davidivadfuldavidivadful
14410
14410
add a comment |
add a comment |
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$begingroup$
Use the equation of a circle with radius $r$ (unknown) and center $(x_0, y_0)$ (unknown).
$endgroup$
– Mauro ALLEGRANZA
3 hours ago
$begingroup$
Then solve te system of equations obtained from the previous one (3 unknown) knowing that points P and Q satisfy the equation of the circle and that the coordinates of the center satisfy the equation of the line.
$endgroup$
– Mauro ALLEGRANZA
3 hours ago