Question

From a point $(h,0)$ common tangents are drawn to circle $x^2+y^2=1$ and $(x-2{)}^2+y^2=4$, value of $h$ is

• A. $2$
• B. $-2$
• C. $-\frac{2}{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is B $-2$

In $△P{T}_1{C}_1$&$△P{T}_2{C}_2$

$\angle C_{1}P{T}_1=\angle C_{2}P{T}_2$ (same angles)

$\angle C_1{T}_{1}P=\angle C_2{T}_{2}P=90^o$

$\therefore \angle P{C}_1{T}_1=\angle P{C}_2{T}_2$ (By angle sum property)

By applying the properties of similar $△s.$

$\frac{P{C}_1}{P{C}_2}=\frac{C_1{T}_1}{C_2{T}_2}$

$\Rightarrow \frac{d_1}{d_2}=\frac{r_1}{r_2}=\frac{1}{2}$

$\Rightarrow 2d_1=d_2$

$\Rightarrow 2h=(h-2)[\because d_1=\sqrt{h^2},d_2={\sqrt{\left(h-2\right)}}^2]$

$\Rightarrow h=-2$

Hence, $h=-2,P(-2,0)$