If the angle between the tangents drawn to the circle $x^2+y^2-12 x-16 y=0$ at the points where the line $5 y=5 x+k$ cut the circle is $60^{\circ}$, then the value of $k$ is

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Aicharya · Accepted Answer

Let the point $P\left(x_1, y_1ight)$, then $ \frac{10}{\sqrt{x_1^2+y_1^2-12 x_1-16 y_1}}=\frac{1}{\sqrt{3}} $ $\Rightarrow \quad x_1^2+y_1^2-12 x_1-16 y_1=300$ Since, line $A B, 5 y=5 x+k$, is a chord of contact of circle with respect to point $P\left(x_1, y_1ight)$, so, $ \begin{aligned} x x_1+y y_1-6\left(x+x_1ight)-8\left(y+y_1ight) & =0 \ x\left(x_1-6ight)+y\left(y_1-8ight)-\left(6 x_1+8 y_1ight) & =0 \ 	herefore \quad \frac{x_1-6}{-5}=\frac{y_1-8}{5}=\frac{6 x_1+8 y_1}{K} & =\lambda(	ext { let }) \end{aligned} $ From Eqs. (i) and (ii), we are getting $ \begin{array}{rlrl} & 25 \lambda^2+25 \lambda^2 =400 \Rightarrow \lambda= \pm 2 \sqrt{2} \ & \lambda=\frac{6 x_1+8 y_1}{k} \ \Rightarrow & K=\frac{6 x_1+8 y_1}{\lambda} =\frac{36-30 \lambda+64+40 \lambda}{\lambda} \ \Rightarrow & k =\frac{100}{\lambda}+10 \Rightarrow k=10 \pm 25 \sqrt{2} \ \Rightarrow & k =5(2 \pm 5 \sqrt{2}) \end{array} $

If the angle between the tangents drawn to the circle $x^2+y^2-12 x-16 y=0$ at the points where the line $5…

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