Question

Let tangents are drawn from $P(3,4)$ to the circle $x^2+y^2=a^2,$ touching the circle at $A$ and $B.$ If area of $△PAB$ is $\frac{192}{25}$ sq. units, then the absolute value of $a$ is

• A. 3.00
• B. 3.0
• C. 3

Answer 1

Answer: (A), (B), (C)

Given circle is, $x^2+y^2=a^2$
Centre, $C\equiv (0,0)$ and radius, $r=|a|$
Equation of chord of contact for point $(3,4)$ is
$x\left(3\right)+y\left(4\right)=a^2$
$\Rightarrow 3x+4y-a^2=0$
Length of tangent from $P$ to the given circle is
$L=\sqrt{S_1}=\sqrt{3^2+4^2-a^2}$
$\Rightarrow L=\sqrt{25-a^2}$ units
Perpendicular distance to the chord of contact from the centre $(0,0)$ of the circle
$p=\frac{|3\times 0-4\times 0-a^2|}{\sqrt{3^2+4^2}}$
$\Rightarrow p=\frac{a^2}{5}$


So, required area of triangle is
$\frac{192}{25}=ar\left(PACB\right)-ar\left(ACB\right)$
$\frac{192}{25}=r\sqrt{S_1}-\left(p\sqrt{r^2-p^2}\right)$
$=|a|\sqrt{25-a^2}-\frac{a^2}{5}\sqrt{a^2-\frac{a^4}{25}}$
$=|a|\sqrt{25-a^2}-\frac{a^2}{25}|a|\sqrt{25-a^2}$
$=|a|\frac{(25-a^2)}{25}\sqrt{25-a^2}$
$=\frac{|a|}{25}(25-a^2{)}^{3/2}$
$\Rightarrow \frac{192}{25}=\frac{|a|(25-a^2{)}^{3/2}}{25}$
$\Rightarrow |a|(25-a^2{)}^{3/2}=3\cdot 2^6\Rightarrow |a|(25-a^2{)}^{3/2}=3\cdot (16{)}^{3/2}=3(25-9{)}^{3/2}$
$\Rightarrow |a|=3$