Question

Two tangents are drawn from $(1,-2)$ to the circle $x^2+y^2-8x+6y+20=0.$ Which of the following is false ?

• A. angle between the tangents is $\frac{\pi }{2}$
• B. equation of one of the two tangents is $x-2y-5=0$
• C. one of the two tangents passes through the origin
• D. sum of the squares of the slopes of the two tangents is $\frac{15}{4}$

Answer 1

The correct option is D sum of the squares of the slopes of the two tangents is $\frac{15}{4}$

For the given circle, $C=(4,-3),r=\sqrt{5}$
Let the equation of any tangent be
$y+2=m(x-1)$
$\Rightarrow mx-y-(m+2)=0...\left(1\right)$

Applying condition for tangency,
$\frac{|4m+3-(m+2)|}{\sqrt{m^2+1}}=\sqrt{5}$
$\Rightarrow (3m+1{)}^2=5(m^2+1)$
$\Rightarrow 2m^2+3m-2=0$
$\Rightarrow m_1=\frac{1}{2},m_2=-2$
($m_1,m_2$ are the slopes of two tangents)

$\Rightarrow m_1\cdot m_2=-1$
$\Rightarrow$ Angle between the tangents is $\frac{\pi }{2}$
Equation of tangents are $x-2y-5=0$ and $2x+y=0$ (from $\left(1\right))$
$2x+y=0$ is passing through the origin.
$\Rightarrow m_1^2+m_2^2=\frac{17}{4}$