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Find equation of the circle whose diameter is the common chord of two other circles?

Find the equation of a circle which is tangent to $y$-axis at a given point and cuts a chord of given length on the $x$-axisFind the length of the chord given that the circle's diameter and the subtended angleCircle Chord Length Given other ChordHow to find the equation of diameter of a circle that passes through the origin?How to find the common chord to $2$ parabolasCircles are drawn through $P$ touching the coordinate axes,such that the length of the common chord of these circles is maximum.Find the ratio $a:b$Find the Length of a Common ChordFind length of chord between two tangents and distance of chord from originGiven two points endpoints of circle chord find the locus of midpointFind chord length given equations for circle and line

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3

1

$begingroup$

Circle 1:

$$x^2 + y^2 +6x + 2y +6 = 0$$

Circle 2:

$$x^2 + y^2 + 8x + y + 10 = 0$$

My attampt:

From circle 1 and 2, I found

$$ y = 2x + 4 $$
which is the common chord.
Pluging that in equation 1 I got

$$5x^2 + 26x + 30 = 0$$

here I got stuck for too much complication if I use quadratic formula.Is there anything wrong?

analytic-geometry circles

share|cite|improve this question

edited 8 hours ago

Quanto

4,95722 silver badges1414 bronze badges

asked 9 hours ago

GhostGhost

5366 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That’s not the common chord: a chord is a line segment. That’s the circles’ radical axis, which is the extension of the common chord.
  $endgroup$
  – amd
  7 hours ago
  

  
  
  
  

add a comment
 | 

3

1

$begingroup$

Circle 1:

$$x^2 + y^2 +6x + 2y +6 = 0$$

Circle 2:

$$x^2 + y^2 + 8x + y + 10 = 0$$

My attampt:

From circle 1 and 2, I found

$$ y = 2x + 4 $$
which is the common chord.
Pluging that in equation 1 I got

$$5x^2 + 26x + 30 = 0$$

here I got stuck for too much complication if I use quadratic formula.Is there anything wrong?

analytic-geometry circles

share|cite|improve this question

edited 8 hours ago

Quanto

4,95722 silver badges1414 bronze badges

asked 9 hours ago

GhostGhost

5366 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That’s not the common chord: a chord is a line segment. That’s the circles’ radical axis, which is the extension of the common chord.
  $endgroup$
  – amd
  7 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

Circle 1:

$$x^2 + y^2 +6x + 2y +6 = 0$$

Circle 2:

$$x^2 + y^2 + 8x + y + 10 = 0$$

My attampt:

From circle 1 and 2, I found

$$ y = 2x + 4 $$
which is the common chord.
Pluging that in equation 1 I got

$$5x^2 + 26x + 30 = 0$$

here I got stuck for too much complication if I use quadratic formula.Is there anything wrong?

analytic-geometry circles

share|cite|improve this question

edited 8 hours ago

Quanto

4,95722 silver badges1414 bronze badges

asked 9 hours ago

GhostGhost

5366 bronze badges

$endgroup$

share|cite|improve this question

edited 8 hours ago

Quanto

4,95722 silver badges1414 bronze badges

asked 9 hours ago

GhostGhost

5366 bronze badges

share|cite|improve this question

edited 8 hours ago

Quanto

4,95722 silver badges1414 bronze badges

asked 9 hours ago

GhostGhost

5366 bronze badges

edited 8 hours ago

Quanto

4,95722 silver badges1414 bronze badges

edited 8 hours ago

Quanto

4,95722 silver badges1414 bronze badges

asked 9 hours ago

GhostGhost

5366 bronze badges

asked 9 hours ago

GhostGhost

5366 bronze badges

3 Answers
3

active

oldest

votes

4

$begingroup$

First, obtain the equations of the intersection points below for both $x$ and $y$,

$$5x^2 + 26x + 30= 0$$
$$5y^2 + 12y -8= 0$$

It may be more efficient not to solve for the intersection points explicitly. Rather, use the relationships from the above equations ,

$$x_1+x_2=-frac{26}{5},>>>x_1x_2=6$$

$$y_1+y_2=-frac{12}{5},>>>y_1y_2=-frac 85 $$

Thus, the center of the circle is $(-13/5, -6/5)$ and its diameter squared is,

$$(x_1-x_2)^2 + (y_1-y_2)^2$$
$$ = (x_1+x_2)^2- 4x_1x_2 + (y_1+y_2)^2 - 4y_1y_2$$

$$= left( frac{26}{5} right)^2 -4cdot 6 + left( frac{12}{5}right)^2 + 4cdot frac 85 = frac{76}{5}$$

The equation of the circle is

$$left( x+frac{13}{5} right)^2 + left( y +frac{6}{5}right)^2 = frac{19}{5}$$

share|cite|improve this answer

edited 4 hours ago

answered 8 hours ago

QuantoQuanto

4,95722 silver badges1414 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

You are on the right track. The numbers cancel out nicely when you sum them. Indeed:
$$5x^2 + 26x + 30 = 0 Rightarrow x_1=frac{-13-sqrt{19}}{5},x_2=frac{-13+sqrt{19}}{5}\
y_1=frac{-6-2sqrt{19}}{5}, y_2=frac{-6+2sqrt{19}}{5}$$
The center of the new circle:
$$frac{x_1+x_2}{2}=-frac{13}{5},frac{y_1+y_2}{2}=-frac{6}{5}$$
The diameter of the new circle:
$$d=sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=sqrt{frac{4cdot 19}{25}+frac{16cdot 19}{25}}=sqrt{frac{76}{5}} Rightarrow \
r=frac12d=sqrt{frac{76}{4cdot 5}}=sqrt{frac{19}{5}}$$
Thus:
$$left(x+frac{13}{5}right)^2+left(y+frac65right)^2=frac{19}{5}.$$

share|cite|improve this answer

answered 8 hours ago

farruhotafarruhota

25.8k22 gold badges1111 silver badges4646 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Comparing this to other solutions, solving the quadratic equation is not so much more complicated after all. Sometimes you just have to not give up so quickly.
  $endgroup$
  – David K
  6 hours ago
  

  
  
  
  

add a comment
 | 

1

$begingroup$

You can certainly keep going the way you are: solve the quadratic equation for $x$ and substitute back into the equation of the radical axis (or either circle equation) to get the endpoints $(x_1,y_1)$ and $(x_2,y_2)$ of the chord. An equation of the circle with that diameter can be written down directly: $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$. Rearrange this into whatever form is required.

However, there’s another way to solve this that requires less work, I think. The circles that pass through the two intersection points form a one-parameter family with equations that can be expressed as an affine combination of the equations of the two circles, to wit, $$(1-lambda)(x^2+y^2+6x+2y+6)+lambda(x^2+y^2+8x+y+10) = 0$$ or $$x^2+y^2+(2lambda+6)x+(2-lambda)y+(4lambda+6)=0.tag{*}$$ By inspection, the coordinates of the center of this circle are $(-lambda-3,lambda/2-1)$ and we want it to lie on the radical axis $y=2x+4$. Substitute the coordinates of the center into this equation and solve the resulting linear equation for $lambda$, then plug that value into equation (*).

share|cite|improve this answer

answered 6 hours ago

amdamd

37.3k22 gold badges1313 silver badges5757 bronze badges

$endgroup$

add a comment
 | 

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4

$begingroup$

First, obtain the equations of the intersection points below for both $x$ and $y$,

$$5x^2 + 26x + 30= 0$$
$$5y^2 + 12y -8= 0$$

It may be more efficient not to solve for the intersection points explicitly. Rather, use the relationships from the above equations ,

$$x_1+x_2=-frac{26}{5},>>>x_1x_2=6$$

$$y_1+y_2=-frac{12}{5},>>>y_1y_2=-frac 85 $$

Thus, the center of the circle is $(-13/5, -6/5)$ and its diameter squared is,

$$(x_1-x_2)^2 + (y_1-y_2)^2$$
$$ = (x_1+x_2)^2- 4x_1x_2 + (y_1+y_2)^2 - 4y_1y_2$$

$$= left( frac{26}{5} right)^2 -4cdot 6 + left( frac{12}{5}right)^2 + 4cdot frac 85 = frac{76}{5}$$

The equation of the circle is

$$left( x+frac{13}{5} right)^2 + left( y +frac{6}{5}right)^2 = frac{19}{5}$$

share|cite|improve this answer

edited 4 hours ago

answered 8 hours ago

QuantoQuanto

4,95722 silver badges1414 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

First, obtain the equations of the intersection points below for both $x$ and $y$,

$$5x^2 + 26x + 30= 0$$
$$5y^2 + 12y -8= 0$$

It may be more efficient not to solve for the intersection points explicitly. Rather, use the relationships from the above equations ,

$$x_1+x_2=-frac{26}{5},>>>x_1x_2=6$$

$$y_1+y_2=-frac{12}{5},>>>y_1y_2=-frac 85 $$

Thus, the center of the circle is $(-13/5, -6/5)$ and its diameter squared is,

$$(x_1-x_2)^2 + (y_1-y_2)^2$$
$$ = (x_1+x_2)^2- 4x_1x_2 + (y_1+y_2)^2 - 4y_1y_2$$

$$= left( frac{26}{5} right)^2 -4cdot 6 + left( frac{12}{5}right)^2 + 4cdot frac 85 = frac{76}{5}$$

The equation of the circle is

$$left( x+frac{13}{5} right)^2 + left( y +frac{6}{5}right)^2 = frac{19}{5}$$

share|cite|improve this answer

edited 4 hours ago

answered 8 hours ago

QuantoQuanto

4,95722 silver badges1414 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

First, obtain the equations of the intersection points below for both $x$ and $y$,

$$5x^2 + 26x + 30= 0$$
$$5y^2 + 12y -8= 0$$

It may be more efficient not to solve for the intersection points explicitly. Rather, use the relationships from the above equations ,

$$x_1+x_2=-frac{26}{5},>>>x_1x_2=6$$

$$y_1+y_2=-frac{12}{5},>>>y_1y_2=-frac 85 $$

Thus, the center of the circle is $(-13/5, -6/5)$ and its diameter squared is,

$$(x_1-x_2)^2 + (y_1-y_2)^2$$
$$ = (x_1+x_2)^2- 4x_1x_2 + (y_1+y_2)^2 - 4y_1y_2$$

$$= left( frac{26}{5} right)^2 -4cdot 6 + left( frac{12}{5}right)^2 + 4cdot frac 85 = frac{76}{5}$$

The equation of the circle is

$$left( x+frac{13}{5} right)^2 + left( y +frac{6}{5}right)^2 = frac{19}{5}$$

share|cite|improve this answer

edited 4 hours ago

answered 8 hours ago

QuantoQuanto

4,95722 silver badges1414 bronze badges

$endgroup$

share|cite|improve this answer

edited 4 hours ago

answered 8 hours ago

QuantoQuanto

4,95722 silver badges1414 bronze badges

answered 8 hours ago

QuantoQuanto

4,95722 silver badges1414 bronze badges

answered 8 hours ago

QuantoQuanto

4,95722 silver badges1414 bronze badges

1

$begingroup$

You are on the right track. The numbers cancel out nicely when you sum them. Indeed:
$$5x^2 + 26x + 30 = 0 Rightarrow x_1=frac{-13-sqrt{19}}{5},x_2=frac{-13+sqrt{19}}{5}\
y_1=frac{-6-2sqrt{19}}{5}, y_2=frac{-6+2sqrt{19}}{5}$$
The center of the new circle:
$$frac{x_1+x_2}{2}=-frac{13}{5},frac{y_1+y_2}{2}=-frac{6}{5}$$
The diameter of the new circle:
$$d=sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=sqrt{frac{4cdot 19}{25}+frac{16cdot 19}{25}}=sqrt{frac{76}{5}} Rightarrow \
r=frac12d=sqrt{frac{76}{4cdot 5}}=sqrt{frac{19}{5}}$$
Thus:
$$left(x+frac{13}{5}right)^2+left(y+frac65right)^2=frac{19}{5}.$$

share|cite|improve this answer

answered 8 hours ago

farruhotafarruhota

25.8k22 gold badges1111 silver badges4646 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Comparing this to other solutions, solving the quadratic equation is not so much more complicated after all. Sometimes you just have to not give up so quickly.
  $endgroup$
  – David K
  6 hours ago
  

  
  
  
  

add a comment
 | 

1

$begingroup$

You are on the right track. The numbers cancel out nicely when you sum them. Indeed:
$$5x^2 + 26x + 30 = 0 Rightarrow x_1=frac{-13-sqrt{19}}{5},x_2=frac{-13+sqrt{19}}{5}\
y_1=frac{-6-2sqrt{19}}{5}, y_2=frac{-6+2sqrt{19}}{5}$$
The center of the new circle:
$$frac{x_1+x_2}{2}=-frac{13}{5},frac{y_1+y_2}{2}=-frac{6}{5}$$
The diameter of the new circle:
$$d=sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=sqrt{frac{4cdot 19}{25}+frac{16cdot 19}{25}}=sqrt{frac{76}{5}} Rightarrow \
r=frac12d=sqrt{frac{76}{4cdot 5}}=sqrt{frac{19}{5}}$$
Thus:
$$left(x+frac{13}{5}right)^2+left(y+frac65right)^2=frac{19}{5}.$$

share|cite|improve this answer

answered 8 hours ago

farruhotafarruhota

25.8k22 gold badges1111 silver badges4646 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Comparing this to other solutions, solving the quadratic equation is not so much more complicated after all. Sometimes you just have to not give up so quickly.
  $endgroup$
  – David K
  6 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

You are on the right track. The numbers cancel out nicely when you sum them. Indeed:
$$5x^2 + 26x + 30 = 0 Rightarrow x_1=frac{-13-sqrt{19}}{5},x_2=frac{-13+sqrt{19}}{5}\
y_1=frac{-6-2sqrt{19}}{5}, y_2=frac{-6+2sqrt{19}}{5}$$
The center of the new circle:
$$frac{x_1+x_2}{2}=-frac{13}{5},frac{y_1+y_2}{2}=-frac{6}{5}$$
The diameter of the new circle:
$$d=sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=sqrt{frac{4cdot 19}{25}+frac{16cdot 19}{25}}=sqrt{frac{76}{5}} Rightarrow \
r=frac12d=sqrt{frac{76}{4cdot 5}}=sqrt{frac{19}{5}}$$
Thus:
$$left(x+frac{13}{5}right)^2+left(y+frac65right)^2=frac{19}{5}.$$

share|cite|improve this answer

answered 8 hours ago

farruhotafarruhota

25.8k22 gold badges1111 silver badges4646 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

farruhotafarruhota

25.8k22 gold badges1111 silver badges4646 bronze badges

answered 8 hours ago

farruhotafarruhota

25.8k22 gold badges1111 silver badges4646 bronze badges

answered 8 hours ago

farruhotafarruhota

25.8k22 gold badges1111 silver badges4646 bronze badges

1

$begingroup$

You can certainly keep going the way you are: solve the quadratic equation for $x$ and substitute back into the equation of the radical axis (or either circle equation) to get the endpoints $(x_1,y_1)$ and $(x_2,y_2)$ of the chord. An equation of the circle with that diameter can be written down directly: $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$. Rearrange this into whatever form is required.

However, there’s another way to solve this that requires less work, I think. The circles that pass through the two intersection points form a one-parameter family with equations that can be expressed as an affine combination of the equations of the two circles, to wit, $$(1-lambda)(x^2+y^2+6x+2y+6)+lambda(x^2+y^2+8x+y+10) = 0$$ or $$x^2+y^2+(2lambda+6)x+(2-lambda)y+(4lambda+6)=0.tag{*}$$ By inspection, the coordinates of the center of this circle are $(-lambda-3,lambda/2-1)$ and we want it to lie on the radical axis $y=2x+4$. Substitute the coordinates of the center into this equation and solve the resulting linear equation for $lambda$, then plug that value into equation (*).

share|cite|improve this answer

answered 6 hours ago

amdamd

37.3k22 gold badges1313 silver badges5757 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

You can certainly keep going the way you are: solve the quadratic equation for $x$ and substitute back into the equation of the radical axis (or either circle equation) to get the endpoints $(x_1,y_1)$ and $(x_2,y_2)$ of the chord. An equation of the circle with that diameter can be written down directly: $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$. Rearrange this into whatever form is required.

However, there’s another way to solve this that requires less work, I think. The circles that pass through the two intersection points form a one-parameter family with equations that can be expressed as an affine combination of the equations of the two circles, to wit, $$(1-lambda)(x^2+y^2+6x+2y+6)+lambda(x^2+y^2+8x+y+10) = 0$$ or $$x^2+y^2+(2lambda+6)x+(2-lambda)y+(4lambda+6)=0.tag{*}$$ By inspection, the coordinates of the center of this circle are $(-lambda-3,lambda/2-1)$ and we want it to lie on the radical axis $y=2x+4$. Substitute the coordinates of the center into this equation and solve the resulting linear equation for $lambda$, then plug that value into equation (*).

share|cite|improve this answer

answered 6 hours ago

amdamd

37.3k22 gold badges1313 silver badges5757 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

You can certainly keep going the way you are: solve the quadratic equation for $x$ and substitute back into the equation of the radical axis (or either circle equation) to get the endpoints $(x_1,y_1)$ and $(x_2,y_2)$ of the chord. An equation of the circle with that diameter can be written down directly: $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$. Rearrange this into whatever form is required.

However, there’s another way to solve this that requires less work, I think. The circles that pass through the two intersection points form a one-parameter family with equations that can be expressed as an affine combination of the equations of the two circles, to wit, $$(1-lambda)(x^2+y^2+6x+2y+6)+lambda(x^2+y^2+8x+y+10) = 0$$ or $$x^2+y^2+(2lambda+6)x+(2-lambda)y+(4lambda+6)=0.tag{*}$$ By inspection, the coordinates of the center of this circle are $(-lambda-3,lambda/2-1)$ and we want it to lie on the radical axis $y=2x+4$. Substitute the coordinates of the center into this equation and solve the resulting linear equation for $lambda$, then plug that value into equation (*).

share|cite|improve this answer

answered 6 hours ago

amdamd

37.3k22 gold badges1313 silver badges5757 bronze badges

$endgroup$

share|cite|improve this answer

answered 6 hours ago

amdamd

37.3k22 gold badges1313 silver badges5757 bronze badges

answered 6 hours ago

amdamd

37.3k22 gold badges1313 silver badges5757 bronze badges

answered 6 hours ago

amdamd

37.3k22 gold badges1313 silver badges5757 bronze badges