Question:
A circle C cuts the circles \(x^{2}+y^{2}-4x+6y+4=0\) and \(x^{2}+y^{2}+6x-4y+9=0\) orthogonally. If origin lies on this circle C, then the radius of the circle C is
A circle C cuts the circles \(x^{2}+y^{2}-4x+6y+4=0\) and \(x^{2}+y^{2}+6x-4y+9=0\) orthogonally. If origin lies on this circle C, then the radius of the circle C is
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For a circle passing through the origin, the constant term \(c\) in the general equation is always zero, which greatly simplifies the orthogonality equations.
Updated On: Jun 9, 2026
- \(\frac{\sqrt{85}}{2}\)
- \(2\sqrt{3}\)
- 4
- 5
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The Correct Option is A
Solution and Explanation
Concept:
A circle \(C\) given by \(x^2 + y^2 + 2gx + 2fy + c = 0\) intersects two other circles \(x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0\) and \(x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0\) orthogonally if it satisfies the orthogonality conditions:
1) \(2g g_1 + 2f f_1 = c + c_1\)
2) \(2g g_2 + 2f f_2 = c + c_2\)
Step 1: Identify the parameters of the given circles.
Circle 1: \(x^2 + y^2 - 4x + 6y + 4 = 0 \implies g_1 = -2, f_1 = 3, c_1 = 4\).
Circle 2: \(x^2 + y^2 + 6x - 4y + 9 = 0 \implies g_2 = 3, f_2 = -2, c_2 = 9\).
Let the circle \(C\) be \(x^2 + y^2 + 2gx + 2fy + c = 0\). Since the origin \((0,0)\) lies on \(C\), we have \(0^2 + 0^2 + 2g(0) + 2f(0) + c = 0\), which implies \(c = 0\).
Step 2: Apply the orthogonality conditions with \(c=0\).
Condition 1: \(2g(-2) + 2f(3) = 0 + 4 \implies -4g + 6f = 4 \implies -2g + 3f = 2\).
Condition 2: \(2g(3) + 2f(-2) = 0 + 9 \implies 6g - 4f = 9\).
Step 3: Solve the system of linear equations for \(g\) and \(f\).
Multiply the first equation by 3: \(-6g + 9f = 6\).
Add to the second equation: \((-6g + 9f) + (6g - 4f) = 6 + 9 \implies 5f = 15 \implies f = 3\).
Substitute \(f=3\) into \(-2g + 3(3) = 2 \implies -2g + 9 = 2 \implies -2g = -7 \implies g = 3.5 = 7/2\).
Step 4: Calculate the radius of circle C.
The radius \(R\) of circle \(C\) is \(\sqrt{g^2 + f^2 - c}\). Since \(c=0\): \[ R = \sqrt{\left(\frac{7}{2}\right)^2 + (3)^2} = \sqrt{\frac{49}{4} + 9} = \sqrt{\frac{49+36}{4}} = \sqrt{\frac{85}{4}} = \frac{\sqrt{85}}{2} \] center minipage0.4
Radius = \(\frac{\sqrt{85}}{2}\) minipage center
Step 1: Identify the parameters of the given circles.
Circle 1: \(x^2 + y^2 - 4x + 6y + 4 = 0 \implies g_1 = -2, f_1 = 3, c_1 = 4\).
Circle 2: \(x^2 + y^2 + 6x - 4y + 9 = 0 \implies g_2 = 3, f_2 = -2, c_2 = 9\).
Let the circle \(C\) be \(x^2 + y^2 + 2gx + 2fy + c = 0\). Since the origin \((0,0)\) lies on \(C\), we have \(0^2 + 0^2 + 2g(0) + 2f(0) + c = 0\), which implies \(c = 0\).
Step 2: Apply the orthogonality conditions with \(c=0\).
Condition 1: \(2g(-2) + 2f(3) = 0 + 4 \implies -4g + 6f = 4 \implies -2g + 3f = 2\).
Condition 2: \(2g(3) + 2f(-2) = 0 + 9 \implies 6g - 4f = 9\).
Step 3: Solve the system of linear equations for \(g\) and \(f\).
Multiply the first equation by 3: \(-6g + 9f = 6\).
Add to the second equation: \((-6g + 9f) + (6g - 4f) = 6 + 9 \implies 5f = 15 \implies f = 3\).
Substitute \(f=3\) into \(-2g + 3(3) = 2 \implies -2g + 9 = 2 \implies -2g = -7 \implies g = 3.5 = 7/2\).
Step 4: Calculate the radius of circle C.
The radius \(R\) of circle \(C\) is \(\sqrt{g^2 + f^2 - c}\). Since \(c=0\): \[ R = \sqrt{\left(\frac{7}{2}\right)^2 + (3)^2} = \sqrt{\frac{49}{4} + 9} = \sqrt{\frac{49+36}{4}} = \sqrt{\frac{85}{4}} = \frac{\sqrt{85}}{2} \] center minipage0.4
Radius = \(\frac{\sqrt{85}}{2}\) minipage center
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