Question:
If a circle \(S\) passes through \((a, b)\) and cuts the circle \(x^2 + y^2 = 4\) orthogonally, then find the locus of the center of \(S\).
If a circle \(S\) passes through \((a, b)\) and cuts the circle \(x^2 + y^2 = 4\) orthogonally, then find the locus of the center of \(S\).
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For a circle \(S\) cutting a circle \(x^2+y^2=R^2\) orthogonally, the power of the center of \(S\) with respect to the circle is \(R^2\).
Updated On: Jun 9, 2026
- \(2ax - 2by + (a^2+b^2+4) = 0\)
- \(2ax + 2by - (a^2+b^2+4) = 0\)
- \(2ax + 2by + (a^2+b^2+4) = 0\)
- \(2ax - 2by - (a^2+b^2+4) = 0\)
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The Correct Option is B
Solution and Explanation
Concept:
Two circles \(x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0\) and \(x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0\) intersect orthogonally if \(2g_1g_2 + 2f_1f_2 = c_1 + c_2\).
Step 1: Assume the equation of circle \(S\) with center \((h, k)\) and radius \(r\).
The circle \(S\) is defined by \((x-h)^2 + (y-k)^2 = r^2\), which expands to: \[ x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0 \] Here, \(g_1 = -h\), \(f_1 = -k\), and \(c_1 = h^2 + k^2 - r^2\).
Step 2: Apply the condition of orthogonality to the circles.
The second circle is \(x^2 + y^2 - 4 = 0\), so \(g_2 = 0\), \(f_2 = 0\), and \(c_2 = -4\). The orthogonality condition is \(2g_1g_2 + 2f_1f_2 = c_1 + c_2\): \[ 2(-h)(0) + 2(-k)(0) = (h^2 + k^2 - r^2) + (-4) \] \[ 0 = h^2 + k^2 - r^2 - 4 \implies r^2 = h^2 + k^2 - 4 \]
Step 3: Utilize the point \((a, b)\) to establish the relationship for the locus.
Since \(S\) passes through \((a, b)\), we have: \[ (a-h)^2 + (b-k)^2 = r^2 \] Substitute the expression for \(r^2\) from Step 2: \[ (a-h)^2 + (b-k)^2 = h^2 + k^2 - 4 \] Expand both sides: \[ a^2 - 2ah + h^2 + b^2 - 2bk + k^2 = h^2 + k^2 - 4 \] \[ a^2 + b^2 - 2ah - 2bk = -4 \] \[ 2ah + 2bk - (a^2 + b^2 + 4) = 0 \]
Step 4: Substitute \(x\) and \(y\) for the coordinates of the center.
Replace \((h, k)\) with \((x, y)\) to find the locus: \[ 2ax + 2by - (a^2 + b^2 + 4) = 0 \] center minipage0.5
Locus: \(2ax + 2by - (a^2+b^2+4) = 0\) minipage center
Step 1: Assume the equation of circle \(S\) with center \((h, k)\) and radius \(r\).
The circle \(S\) is defined by \((x-h)^2 + (y-k)^2 = r^2\), which expands to: \[ x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0 \] Here, \(g_1 = -h\), \(f_1 = -k\), and \(c_1 = h^2 + k^2 - r^2\).
Step 2: Apply the condition of orthogonality to the circles.
The second circle is \(x^2 + y^2 - 4 = 0\), so \(g_2 = 0\), \(f_2 = 0\), and \(c_2 = -4\). The orthogonality condition is \(2g_1g_2 + 2f_1f_2 = c_1 + c_2\): \[ 2(-h)(0) + 2(-k)(0) = (h^2 + k^2 - r^2) + (-4) \] \[ 0 = h^2 + k^2 - r^2 - 4 \implies r^2 = h^2 + k^2 - 4 \]
Step 3: Utilize the point \((a, b)\) to establish the relationship for the locus.
Since \(S\) passes through \((a, b)\), we have: \[ (a-h)^2 + (b-k)^2 = r^2 \] Substitute the expression for \(r^2\) from Step 2: \[ (a-h)^2 + (b-k)^2 = h^2 + k^2 - 4 \] Expand both sides: \[ a^2 - 2ah + h^2 + b^2 - 2bk + k^2 = h^2 + k^2 - 4 \] \[ a^2 + b^2 - 2ah - 2bk = -4 \] \[ 2ah + 2bk - (a^2 + b^2 + 4) = 0 \]
Step 4: Substitute \(x\) and \(y\) for the coordinates of the center.
Replace \((h, k)\) with \((x, y)\) to find the locus: \[ 2ax + 2by - (a^2 + b^2 + 4) = 0 \] center minipage0.5
Locus: \(2ax + 2by - (a^2+b^2+4) = 0\) minipage center
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