Question

Point to the straight line $x-y+1=0$, the tangents from which to the circle $x^2+y^2-3x=0$ are of length $2unit$ is

• A. $\frac{-3}{2},52$
• B. $\frac{-3}{2},-52$
• C. $\frac{3}{2},52$
• D. None of these

Answer 1

The correct option is D None of these

Let the equation of the circle be represented as $S(x,y)=0.$

If the point in question is (h,k),the length of the tangents drawn to the circle from this point are given by $\sqrt{S(h,k)}.$

So,$\sqrt{h^2+k^2-3h}=2$

$h^2+k^2-3h-4=0$

$(h^2-3h+\frac{9}{4})+k^2=4+\frac{9}{4}$

${(h-\frac{3}{2})}^2+k^2=52$

This is one locus.

The required point lies on the line:$x-y+1=0.$

So,$h-k+1=0.$

Substituting $k=h+1$,and solving $f$ or $h$,then finding $k$,we get the two possible points as $(-1,0)$ and $(\frac{3}{2},\frac{5}{2})$.