Question

If the equation of normal to the circle $x^2+y^2-16x-12y+99=0$, which is also a tangent to $x^2+y^2=4$ is ax + by + c = 0, find the value of $\left[|\right(\frac{a}{b}\left)|\right]$, where [x] is the greatest integer function

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (A), (B)

The correct options are
A

0

B

1

Any normal to $c_1$ passes through centre (8,6).Let the equation of tangent be y=mx+c

$\Rightarrow c=\pm 2\sqrt{a^2+b^2}$

We have 6=8m+c

$\Rightarrow c=6-8m$ .........(1)

Perpendicular distance of mx-y+c=0 from centre of $c_2$ is equal to its radius

$\Rightarrow \left|\frac{c}{\sqrt{1+m^2}}\right|=2$

Using (1) $\Rightarrow \left|\frac{6-8m}{\sqrt{1+m^2}}\right|=2$

$\Rightarrow (8m-6{)}^2=4(1+m^2)$

$\Rightarrow 64m^2+36-96m=4+4m^2$

$\Rightarrow 60m^2-96m+32=0$

$\Rightarrow 15m^2-24m+8=0$

$m=\frac{24\pm \sqrt{{24}^2-4\times 15\times 8}}{2\times 15}$

$=\frac{24\pm \sqrt{96}}{2\times 30}=\frac{24\pm 4\sqrt{6}}{30}$

We are given the equation of tangent as ax+by+c=0. The slope of this line is $\frac{-a}{b}$

Which found as $m=\frac{24\pm 4\sqrt{6}}{30}$. Now we want to find $\left[\left|\frac{a}{b}\right|\right]$,which is nothing but [|m|] .

We will get two values for this .

For $m=\frac{24+4\sqrt{6}}{30}$

[|m|] =1

If $m=\frac{24-4\sqrt{6}}{30}$

[|m|] =0