Question:
The radical centre of the circles
\[
x^2+y^2+3x+2y+1=0,
\]
\[
x^2+y^2-x+6y+5=0
\]
and
\[
x^2+y^2+5x-8y+15=0
\]
is
The radical centre of the circles
\[
x^2+y^2+3x+2y+1=0,
\]
\[
x^2+y^2-x+6y+5=0
\]
and
\[
x^2+y^2+5x-8y+15=0
\]
is
Show Hint
To find the radical centre of three circles:
(i) Find two radical axes by subtraction
and
(ii) Solve them simultaneously.
Updated On: Jun 26, 2026
- \((3,2)\)
- \((-3,-2)\)
- \((2,3)\)
- \((-2,-3)\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Recall the meaning of radical centre.
The radical centre is the common point of intersection of the radical axes of the circles.
We obtain the radical axis by subtracting equations of circles pairwise.
Step 2: Find the radical axis of first and second circles.
Subtract \[ x^2+y^2-x+6y+5=0 \] from \[ x^2+y^2+3x+2y+1=0 \] We get \[ 3x+2y+1-(-x+6y+5)=0 \] \[ 3x+2y+1+x-6y-5=0 \] \[ 4x-4y-4=0 \] \[ x-y-1=0 \]
Step 3: Find the radical axis of second and third circles.
Subtract \[ x^2+y^2+5x-8y+15=0 \] from \[ x^2+y^2-x+6y+5=0 \] We get \[ -x+6y+5-(5x-8y+15)=0 \] \[ -x+6y+5-5x+8y-15=0 \] \[ -6x+14y-10=0 \] \[ 3x-7y+5=0 \]
Step 4: Solve the two radical axes simultaneously.
The equations are \[ x-y-1=0 \] and \[ 3x-7y+5=0 \] From the first equation, \[ x=y+1 \] Substitute in the second equation: \[ 3(y+1)-7y+5=0 \] \[ 3y+3-7y+5=0 \] \[ -4y+8=0 \] \[ y=2 \]
Step 5: Find \(x\).
Using \[ x=y+1, \] we get \[ x=2+1=3 \]
Step 6: Write the radical centre.
Thus, the common point is \[ (3,2) \]
Step 7: Final conclusion.
Therefore, the radical centre is \[ \boxed{(3,2)} \]
The radical centre is the common point of intersection of the radical axes of the circles.
We obtain the radical axis by subtracting equations of circles pairwise.
Step 2: Find the radical axis of first and second circles.
Subtract \[ x^2+y^2-x+6y+5=0 \] from \[ x^2+y^2+3x+2y+1=0 \] We get \[ 3x+2y+1-(-x+6y+5)=0 \] \[ 3x+2y+1+x-6y-5=0 \] \[ 4x-4y-4=0 \] \[ x-y-1=0 \]
Step 3: Find the radical axis of second and third circles.
Subtract \[ x^2+y^2+5x-8y+15=0 \] from \[ x^2+y^2-x+6y+5=0 \] We get \[ -x+6y+5-(5x-8y+15)=0 \] \[ -x+6y+5-5x+8y-15=0 \] \[ -6x+14y-10=0 \] \[ 3x-7y+5=0 \]
Step 4: Solve the two radical axes simultaneously.
The equations are \[ x-y-1=0 \] and \[ 3x-7y+5=0 \] From the first equation, \[ x=y+1 \] Substitute in the second equation: \[ 3(y+1)-7y+5=0 \] \[ 3y+3-7y+5=0 \] \[ -4y+8=0 \] \[ y=2 \]
Step 5: Find \(x\).
Using \[ x=y+1, \] we get \[ x=2+1=3 \]
Step 6: Write the radical centre.
Thus, the common point is \[ (3,2) \]
Step 7: Final conclusion.
Therefore, the radical centre is \[ \boxed{(3,2)} \]
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