Question

The chord of contact of tangents from a point P to a circle passes through Q. If $l_1$ and $l_2$ are the lengths of the tangents from P and Q to the circle, then PQ is equal to

• A. $\frac{l_1+l_2}{2}$
• B. $\frac{l_1-l_2}{2}$
• C. $\sqrt{l_1^2+l_2^2}$
• D. $2\sqrt{l_1^2+l_2^2}$

Answer 1

The correct option is C $\sqrt{l_1^2+l_2^2}$

Let $P(x_1,y_1)$ and $Q(x_2,y_2)$ be two points and $x^2+y^2=a^2$ be the given circle. Then, the chord of contact of tangents drawn from P to the given circle is $x{x}_1+y{y}_1=a^2$.
It will pass through $Q(x_2,y_2)$, if
$x_1{x}_2+x_1{y}_2=a^2$ ......... (i)
Now, $l_1=\sqrt{x_1^2+y_1^2-a^2}$,
$l_2=\sqrt{x_2^2+y_2^2-a^2}$
and $PQ=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
$=\sqrt{(x_2^2+y_2^2)+(x_1^2+y_1^2)-2(x_1{x}_2+y_1{y}_2)}$
$\therefore PQ=\sqrt{\left[\right(x_2^2+y_2^2)+(x_1^2+y_1^2)-2a^2]}$ [Using equation (i)]
$\Rightarrow PQ=\sqrt{(x_1^2+y_1^2-a^2)+(x_2^2+y_2^2-a^2)}$
$\Rightarrow PQ=\sqrt{l_1^2+l_2^2}$