Question

If $a^2-b{m}^2+2d+1=0$, where $a,b,d$ are fixed real numbers such that $a+b=d^2$, then the line $x+my+1=0$ touches a fixed circle

• A. which cuts the y-axis orthogonally
• B. with radius equal to $b$
• C. on which the length of the tangent from the origin is $\sqrt{a^2-b}$
• D. none of these

Answer 1

The correct option is C on which the length of the tangent from the origin is $\sqrt{a^2-b}$

Given, $a^2-b{m}^2+2d+1=0a,b,c,d\in R$ touching line $x+my+1=0$

If the line touches the circle then,

radius $=$distance of line from centre.

Let centre of circle be $(h,k)$ then the distance from line is

$\frac{|h+mk+1|}{\sqrt{1+m^2}}=r{h}^2+m^2{k}^2+1+2hmk+2mk+2h=r^2+r^2{m}^2{h}^2+m^2(k^2-r^2)+2hmk+2mk+2h+1-r^2=0$

Comparing it with the given :

$a=h,r^2=b,k=0,2a-b=2d$

Equation of circle $(x-a{)}^2+y^2=b$

Length of tangent from origin $S=\sqrt{a^2+0-b}=\sqrt{a^2-b}$