Question

Let the tangent to the circle $x^2+y^2=25$ at the point $R(3,4)$ meet the $x$-axis and $y$-axis at points $P\&Q$, respectively. If $r$ is the radius of the circle passing through the origin $O$ and having a centre at the incentre of the triangle $OPQ$, then$r^2$ is equal to:

• A. $\frac{625}{72}$
• B. $\frac{585}{66}$
• C. $\frac{125}{72}$
• D. $\frac{529}{64}$

Answer 1

The correct option is A

$\frac{625}{72}$

Explanation for the correct option:

Finding the value of $r^2$

Given the equation of the circle,$x^2+y^2=25$

We know that the equation of a tangent to the circle is $x{x}_1+y{y}_1=c$

And the given point is $R(3,4)$$=(x_1,y_1)$

Then equation of tangent to this circle,

$\Rightarrow 3x+4y=25$

Now putting $x=0\&y=0$in this equation respectively we get,

$y=\frac{25}{4}\&x=\frac{25}{3}$

Now illustration of figure,

So, incentre $=\left(\frac{{\displaystyle \frac{25}{4}}\times \frac{25}{3}}{25},\frac{\frac{25}{4}\times \frac{25}{3}}{25}\right)$ $$\begin{gathered} (\frac{{\mathrm{ax}}_1+{\mathrm{bx}}_2+{\mathrm{cx}}_3}{a+b+c},\frac{{\mathrm{ay}}_1+{\mathrm{by}}_2+{\mathrm{cy}}_3}{a+b+c}), \\ \left(a,b,c\mathrm{are}\mathrm{sides}(x_k,y_k)\mathrm{are}\mathrm{vertices}\right) \end{gathered}$$

$=\left(\frac{25}{12},\frac{25}{12}\right)$

Radius will be equal to distance of incentre fro origin

$$\begin{gathered} \therefore r^2=2\times {\left(\frac{25}{12}\right)}^2 \\ =\frac{625}{72} \end{gathered}$$

Hence, the correct answer is option (A)