Question:
The centre of the circle which intersects the circles
\[
x^2+y^2-8x+10y+5=0
\]
and
\[
x^2+y^2-2x+2y+1=0
\]
orthogonally is:
The centre of the circle which intersects the circles
\[
x^2+y^2-8x+10y+5=0
\]
and
\[
x^2+y^2-2x+2y+1=0
\]
orthogonally is:
Show Hint
For two orthogonal circles,
\[
2gg_1+2ff_1=c+c_1.
\]
This condition is one of the most important results in coordinate geometry and frequently appears in competitive examinations.
Updated On: Jun 10, 2026
- \((-6,-4)\)
- \((6,4)\)
- \((3,5)\)
- \((-3,-5)\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
Two circles intersect orthogonally if the tangents at their points of intersection are perpendicular. A very important condition for orthogonal intersection of two circles is
\[
2gg_1+2ff_1=c+c_1,
\]
where
\[
x^2+y^2+2gx+2fy+c=0
\]
and
\[
x^2+y^2+2g_1x+2f_1y+c_1=0
\]
are the equations of the circles.
In this problem, we are asked to find a circle that cuts two given circles orthogonally. Therefore, the unknown circle must satisfy the orthogonality condition with both circles simultaneously.
Step 1: Write the general equation of the required circle. Let the required circle be \[ x^2+y^2+2gx+2fy+c=0. \] Its centre is \[ (-g,-f). \] Our objective is to determine \(g\) and \(f\).
Step 2: Compare the first given circle with the standard form. Given \[ x^2+y^2-8x+10y+5=0. \] Comparing, \[ g_1=-4,\qquad f_1=5,\qquad c_1=5. \] Applying orthogonality, \[ 2g(-4)+2f(5)=c+5. \] Hence, \[ -8g+10f=c+5. \] \[ c=-8g+10f-5. \]
Step 3: Compare the second given circle. Given \[ x^2+y^2-2x+2y+1=0. \] Comparing, \[ g_2=-1,\qquad f_2=1,\qquad c_2=1. \] Using orthogonality again, \[ 2g(-1)+2f(1)=c+1. \] Thus, \[ -2g+2f=c+1. \] \[ c=-2g+2f-1. \]
Step 4: Equate both expressions of \(c\). \[ -8g+10f-5 = -2g+2f-1. \] Simplifying, \[ -6g+8f=4. \] Dividing by \(2\), \[ -3g+4f=2. \]
Step 5: Check the options. The centre is \[ (-g,-f). \] Testing option \((6,4)\), \[ g=-6,\qquad f=-4. \] Substituting, \[ -3(-6)+4(-4) = 18-16 = 2. \] The relation is satisfied. Hence the centre is \[ (6,4). \]
Step 6: Final Conclusion. Therefore, the centre of the required circle is \[ \boxed{(6,4)}. \] Hence the correct answer is \[ \boxed{\text{Option (B)}}. \]
Step 1: Write the general equation of the required circle. Let the required circle be \[ x^2+y^2+2gx+2fy+c=0. \] Its centre is \[ (-g,-f). \] Our objective is to determine \(g\) and \(f\).
Step 2: Compare the first given circle with the standard form. Given \[ x^2+y^2-8x+10y+5=0. \] Comparing, \[ g_1=-4,\qquad f_1=5,\qquad c_1=5. \] Applying orthogonality, \[ 2g(-4)+2f(5)=c+5. \] Hence, \[ -8g+10f=c+5. \] \[ c=-8g+10f-5. \]
Step 3: Compare the second given circle. Given \[ x^2+y^2-2x+2y+1=0. \] Comparing, \[ g_2=-1,\qquad f_2=1,\qquad c_2=1. \] Using orthogonality again, \[ 2g(-1)+2f(1)=c+1. \] Thus, \[ -2g+2f=c+1. \] \[ c=-2g+2f-1. \]
Step 4: Equate both expressions of \(c\). \[ -8g+10f-5 = -2g+2f-1. \] Simplifying, \[ -6g+8f=4. \] Dividing by \(2\), \[ -3g+4f=2. \]
Step 5: Check the options. The centre is \[ (-g,-f). \] Testing option \((6,4)\), \[ g=-6,\qquad f=-4. \] Substituting, \[ -3(-6)+4(-4) = 18-16 = 2. \] The relation is satisfied. Hence the centre is \[ (6,4). \]
Step 6: Final Conclusion. Therefore, the centre of the required circle is \[ \boxed{(6,4)}. \] Hence the correct answer is \[ \boxed{\text{Option (B)}}. \]
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