Question

If the lengths of the tangents drawn from the point $(1,2)$ to the circle $x^2+y^2+x+y-4=0$ and $3x^2+3y^2-x-y+k=0$ be in the ratio $4:3$, then $k=$

• A. $6$
• B. $7$
• C. $\frac{-21}{4}$
• D. $8$

Answer 1

Answer: (C)

$\frac{-21}{4}$

Let S1 and S2 be the two circles given.
S1 us given by S1:$x^2+y^2+x+y-4=0$
S2 is given by S2:$x^2+y^2-\frac{x}{3}-\frac{y}{3}-k=0$$x^2+y^2+x+y-4=0$
(1,2) is the point from where tangents are drawn to the two circles S1 and S2.
Length of a tangent drawn from a point(x1,y1) to a circle $x^2+y^2+2gx+2fy+c=0$ is given by:
Length of tangent,L=$\sqrt{x{1}^2+y{1}^2+2gx1+2fy1+c}\to \left(1\right)$
Let L1 and L2be the length of the tangents drawn from (1,2).
given,$\frac{L1}{L2}=\frac{4}{3}$

i.e $\frac{\sqrt{1^2+2^2+1+2-4}}{\sqrt{1^2+2^2-\frac{1}{3}-\frac{3}{3}-\frac{k}{3}}}=\frac{4}{3}$

simplifying and squaring on both sides:
$\frac{4}{4+\frac{k}{3}}=\frac{16}{9}$

cancelling common terms
$\frac{1}{4+\frac{k}{3}}=\frac{4}{9}$

cross multiplying
$9=16+\frac{4k}{3}$

$\Rightarrow k=\frac{-21}{4}$

Thereofore option (c) is the correct answer