Question

If the lengths of the tangents from the point (1,2) to the circles $ {{x}^{2}}+{{y}^{2}}+x+y-4=0 $ and $ 3{{x}^{2}}+3{{y}^{2}}-x-y-\lambda =0 $ are in the ratio $ 4:3, $ then the value of $ \lambda $ is

• A. $ \frac{21}{2} $
• B. $ \frac{21}{4} $
• C. $ \frac{21}{5} $
• D. $ \frac{21}{11} $

Answer 1

Answer: (B)

Length of tangent from the point (1, 2) to the circle $ {{x}^{2}}+{{y}^{2}}+x+y-4=0 $ is $ \sqrt{1+4+1+2-4}=2 $ Similarly, length of tangent from the point (1,2) to the circle $ 3{{x}^{2}}+3{{y}^{2}}-x-y-\lambda =0 $ or $ {{x}^{2}}+{{y}^{2}}-\frac{x}{3}-\frac{y}{3}-\frac{\lambda }{3}=0 $ is $ \sqrt{1+4-\frac{1}{3}-\frac{2}{3}-\frac{\lambda }{3}}=\sqrt{4-\frac{\lambda }{3}} $ But given that ratio of lengths of tangents to two circles is $ 4:3 $ . $ \therefore $ $ \frac{2}{\sqrt{4-\frac{\lambda }{3}}}=\frac{4}{3} $ $ \Rightarrow $ $ 2\sqrt{4-\frac{\lambda }{3}}=3 $ On squaring, $ 4\left( 4-\frac{\lambda }{3} \right)=9 $ $ \Rightarrow $ $ 4-\frac{\lambda }{3}=\frac{9}{4} $ $ \Rightarrow $ $ \frac{\lambda }{3}=4-\frac{9}{7}=\frac{16-9}{4}=\frac{7}{4} $ $ \Rightarrow $ $ \lambda =\frac{21}{4} $