Search any question & find its solution
Question:
Answered & Verified by Expert
The radius of the circle $r=12 \cos \theta+5 \sin \theta$ is
Options:
Solution:
1291 Upvotes
Verified Answer
The correct answer is:
$\frac{13}{2}$
Given equation of circle is
$r=12 \cos \theta+5 \sin \theta$
Put $\cos \theta=\frac{x}{r}$ and $\sin \theta=\frac{y}{r}$, we get
$\begin{array}{rlrl} & & r & =12 \times \frac{x}{r}+5 \times \frac{y}{r} \\ & \Rightarrow & r^2 & =12 x+5 y \\ \Rightarrow & x^2+y^2 & =12 x+5 y \\ \Rightarrow & x^2+y^2-12 x-5 y & =0\end{array}$
$\therefore$ Centre is $\left(6, \frac{5}{2}\right)$
$\begin{aligned} \therefore \text { Radius of circle } & =\sqrt{(6)^2+\left(\frac{5}{2}\right)^2} \\ & =\sqrt{36+\frac{25}{4}}=\sqrt{\frac{144+25}{4}} \\ & =\sqrt{\frac{169}{4}}=\frac{13}{2}\end{aligned}$
$r=12 \cos \theta+5 \sin \theta$
Put $\cos \theta=\frac{x}{r}$ and $\sin \theta=\frac{y}{r}$, we get
$\begin{array}{rlrl} & & r & =12 \times \frac{x}{r}+5 \times \frac{y}{r} \\ & \Rightarrow & r^2 & =12 x+5 y \\ \Rightarrow & x^2+y^2 & =12 x+5 y \\ \Rightarrow & x^2+y^2-12 x-5 y & =0\end{array}$
$\therefore$ Centre is $\left(6, \frac{5}{2}\right)$
$\begin{aligned} \therefore \text { Radius of circle } & =\sqrt{(6)^2+\left(\frac{5}{2}\right)^2} \\ & =\sqrt{36+\frac{25}{4}}=\sqrt{\frac{144+25}{4}} \\ & =\sqrt{\frac{169}{4}}=\frac{13}{2}\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.