Question

If \( \left(\frac{1}{10}, \frac{-1}{5}\right) \) is the inverse point of a point (-1, 2) with respect to the circle \( x^2+y^2-2x+4y+c=0 \) then c =

• A. 4
• B. -4
• C. 2
• D. -2

Hint:

If Q is the inverse point of P w.r.t. a circle (centre C, radius R), then C,P,Q are collinear and \( CP \cdot CQ = R^2 \). Calculate the centre C and radius squared \(R^2\) in terms of \(c\). Calculate distances CP and CQ. Use the property to solve for \(c\).

Answer 1

The Correct Option is B

Let \( P=(-1,2) \) and its inverse point \( Q = \left(\frac{1}{10}, \frac{-1}{5}\right) \).
Circle \( S \equiv x^2+y^2-2x+4y+c=0 \).
Centre \( C = (-g,-f) \).
Here \( 2g=-2 \implies g=-1 \); \( 2f=4 \implies f=2 \).
So, centre \( C = (1,-2) \).
Radius \( R = \sqrt{g^2+f^2-c} = \sqrt{(-1)^2+2^2-c} = \sqrt{1+4-c} = \sqrt{5-c} \).
Property of inverse points: C, P, Q are collinear and \( CP \cdot CQ = R^2 \).
Distance CP: \[ CP = \sqrt{(-1-1)^2 + (2-(-2))^2} = \sqrt{(-2)^2 + 4^2} = \sqrt{4+16} = \sqrt{20} \] Distance CQ: \[ CQ = \sqrt{\left(\frac{1}{10}-1\right)^2 + \left(\frac{-1}{5}-(-2)\right)^2} = \sqrt{\left(-\frac{9}{10}\right)^2 + \left(\frac{9}{5}\right)^2} \] \[ = \sqrt{\frac{81}{100} + \frac{81}{25}} = \sqrt{\frac{81}{100} + \frac{324}{100}} = \sqrt{\frac{405}{100}} = \sqrt{\frac{81 \times 5}{100}} = \frac{9\sqrt{5}}{10} \] Using \( CP \cdot CQ = R^2 \): \[ \sqrt{20} \cdot \frac{9\sqrt{5}}{10} = R^2 \] \[ (2\sqrt{5}) \cdot \frac{9\sqrt{5}}{10} = R^2 \] \[ \frac{18 \cdot 5}{10} = R^2 \implies \frac{90}{10} = R^2 \implies 9 = R^2 \] We have \( R^2 = 5-c \).
\[ 9 = 5-c \implies c = 5-9 = -4 \] This matches option (2).