Question:
The radical centre of the three circles
\[
x^2+y^2-1=0,
\]
\[
x^2+y^2-8x+15=0
\]
and
\[
x^2+y^2+10y+24=0
\]
is
The radical centre of the three circles
\[
x^2+y^2-1=0,
\]
\[
x^2+y^2-8x+15=0
\]
and
\[
x^2+y^2+10y+24=0
\]
is
Show Hint
The radical centre of three circles is obtained by finding the intersection point of any two radical axes.
Updated On: Jun 22, 2026
- \(\left(2,-\frac52\right)\)
- \(\left(2,\frac52\right)\)
- \(\left(-2,\frac52\right)\)
- \(\left(-2,-\frac52\right)\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Write the three circles.
Let \[ S_1=x^2+y^2-1=0 \] \[ S_2=x^2+y^2-8x+15=0 \] \[ S_3=x^2+y^2+10y+24=0 \]
Step 2: Find radical axis of \(S_1\) and \(S_2\).
Subtract \(S_1\) from \(S_2\): \[ (x^2+y^2-8x+15)-(x^2+y^2-1)=0 \] \[ -8x+16=0 \] \[ x=2 \]
Step 3: Find radical axis of \(S_1\) and \(S_3\).
Subtract \(S_1\) from \(S_3\): \[ (x^2+y^2+10y+24)-(x^2+y^2-1)=0 \] \[ 10y+25=0 \] \[ y=-\frac52 \]
Step 4: Find radical centre.
The radical centre is the point of intersection of radical axes.
So, \[ x=2,\qquad y=-\frac52 \]
Step 5: Final conclusion.
Therefore, the radical centre is \[ \boxed{\left(2,-\frac52\right)} \]
Let \[ S_1=x^2+y^2-1=0 \] \[ S_2=x^2+y^2-8x+15=0 \] \[ S_3=x^2+y^2+10y+24=0 \]
Step 2: Find radical axis of \(S_1\) and \(S_2\).
Subtract \(S_1\) from \(S_2\): \[ (x^2+y^2-8x+15)-(x^2+y^2-1)=0 \] \[ -8x+16=0 \] \[ x=2 \]
Step 3: Find radical axis of \(S_1\) and \(S_3\).
Subtract \(S_1\) from \(S_3\): \[ (x^2+y^2+10y+24)-(x^2+y^2-1)=0 \] \[ 10y+25=0 \] \[ y=-\frac52 \]
Step 4: Find radical centre.
The radical centre is the point of intersection of radical axes.
So, \[ x=2,\qquad y=-\frac52 \]
Step 5: Final conclusion.
Therefore, the radical centre is \[ \boxed{\left(2,-\frac52\right)} \]
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