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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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3

$begingroup$

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

share|cite|improve this question

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 | 

3

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

edited 2 hours ago

dantopa

6,74442345

asked 3 hours ago

injustice fellowinjustice fellow

312

$endgroup$

share|cite|improve this question

edited 2 hours ago

dantopa

6,74442345

asked 3 hours ago

injustice fellowinjustice fellow

312

share|cite|improve this question

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

asked 3 hours ago

injustice fellowinjustice fellow

312

asked 3 hours ago

injustice fellowinjustice fellow

312

5 Answers
5

active

oldest

votes

4

$begingroup$

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

share|cite|improve this answer

edited 2 hours ago

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 | 

1

$begingroup$

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...

share|cite|improve this answer

answered 2 hours ago

Maria MazurMaria Mazur

50.7k1362126

$endgroup$

add a comment | 

1

$begingroup$

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 $5$.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Yves DaoustYves Daoust

134k676232

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  (+).................
  $endgroup$
  – E.H.E
  2 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

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$$(x+3)^2+(y-1)^2 = 25$$

share|cite|improve this answer

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 | 

0

$begingroup$

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.

share|cite|improve this answer

answered 2 hours ago

davidivadfuldavidivadful

14410

$endgroup$

add a comment | 

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4

$begingroup$

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

share|cite|improve this answer

edited 2 hours ago

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 | 

4

$begingroup$

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

share|cite|improve this answer

edited 2 hours ago

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

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

EagleEagle

19313

$endgroup$

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

EagleEagle

19313

answered 2 hours ago

EagleEagle

19313

answered 2 hours ago

EagleEagle

19313

1

$begingroup$

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...

share|cite|improve this answer

answered 2 hours ago

Maria MazurMaria Mazur

50.7k1362126

$endgroup$

add a comment | 

1

$begingroup$

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...

share|cite|improve this answer

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...

share|cite|improve this answer

answered 2 hours ago

Maria MazurMaria Mazur

50.7k1362126

$endgroup$

share|cite|improve this answer

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

1

$begingroup$

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 $5$.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Yves DaoustYves Daoust

134k676232

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  (+).................
  $endgroup$
  – E.H.E
  2 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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 $5$.

share|cite|improve this answer

edited 1 hour ago

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 $5$.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Yves DaoustYves Daoust

134k676232

$endgroup$

share|cite|improve this answer

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

0

$begingroup$

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$$(x+3)^2+(y-1)^2 = 25$$

share|cite|improve this answer

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 | 

0

$begingroup$

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$$(x+3)^2+(y-1)^2 = 25$$

share|cite|improve this answer

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$$(x+3)^2+(y-1)^2 = 25$$

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answered 2 hours ago

Ak19Ak19

1425

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$endgroup$

share|cite|improve this answer

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

0

$begingroup$

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.

share|cite|improve this answer

answered 2 hours ago

davidivadfuldavidivadful

14410

$endgroup$

add a comment | 

0

$begingroup$

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.

share|cite|improve this answer

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.

share|cite|improve this answer

answered 2 hours ago

davidivadfuldavidivadful

14410

$endgroup$

share|cite|improve this answer

answered 2 hours ago

davidivadfuldavidivadful

14410

answered 2 hours ago

davidivadfuldavidivadful

14410

answered 2 hours ago

davidivadfuldavidivadful

14410