Question

The locus of a point from which the lengths of the tangents to the circles $x^2+y^2=4$ and $2(x^2+y^2)-10x+3y-2=0$ are equal to

• A. a straight line inclined at $\pi /4$ with the line joining the centers of the circles
• B. a circle
• C. an ellipse
• D. a straight line perpendicular to the line joining the centers of the circles

Answer 1

Answer: (D)

a straight line perpendicular to the line joining the centers of the circles

Given circles may be written as,

$x^2+y^2=2^2$ and ${(x-\frac{5}{2})}^2+{(y+\frac{3}{4})}^2={\left(\frac{5\sqrt{5}}{4}\right)}^2$

$\Rightarrow r_1=2,C_1=(0,0),r_2=\frac{5\sqrt{5}}{4},C_2=(\frac{5}{2},-\frac{3}{4})$

And let the point from which tangent is drawn to both the circle be $P(h,k)$
Clearly $△P{T}_1{C}_1$ and $\Delta P{T}_2{C}_2$ are right angled at $T_1$ and $T_2$ respectively.

$\Rightarrow (P{T}_1{)}^2=(P{C}_1{)}^2-r_1^2=h^2+k^2-4$

and $(P{T}_2{)}^2=(P{C}_2{)}^2-r_2^2={(h-\frac{5}{2})}^2+{(k+\frac{3}{4})}^2-{\left(\frac{5\sqrt{5}}{4}\right)}^2=h^2+k^2-5h+3/2k-1$

Now using given condition $P{T}_1=P{T}_2$

$\Rightarrow (P{T}_1{)}^2=(P{T}_2{)}^2$

$\Rightarrow 5h-3/2k=3$

$\Rightarrow 10h-3k=6$

Hence, required locus of $P(h,k)$ is $10x-3y=6$

Clearly this a straight line perpendicular to the line joining $C_1{C}_2$.