Question

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

• A. $(-\frac{3}{2},\frac{5}{2})$
• B. $(\frac{-3}{2},-\frac{5}{2})$
• C. $(\frac{3}{2},\frac{5}{2})$
• D. $(\frac{3}{2},-\frac{5}{2})$

Answer 1

Answer: (C)

$(\frac{3}{2},\frac{5}{2})$

R.E.F image

$x^2+y^2-3x=0\equiv 5$

length of tangent from point

$P(\alpha ,\alpha +1)$

$=\sqrt{S_1}=2$ (given)

$=\sqrt{{\alpha }^2+(\alpha +1{)}^2-3\alpha }=2$

$\Rightarrow {\alpha }^2+{\alpha }^2+2\alpha +1-3\alpha =4$

$2{\alpha }^2-\alpha -3=0$

$\alpha =\frac{+1\pm \sqrt{1+24}}{4}$

$\alpha =\frac{1\pm 5}{4}$

$\alpha =\frac{3}{2},-1$

So required Point is

$(\alpha ,\alpha +1),\equiv (\frac{3}{2},\frac{5}{2})$

Option (C)