Search any question & find its solution
Question:
Answered & Verified by Expert
The radius of the circle $x^{2}+y^{2}+x+c=0$ passing through the origin is
Options:
Solution:
2285 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2}$
Circle is passing through origin then $\mathrm{C}=0$ Now, $x^{2}+y^{2}+x=0$
$\mathrm{x}^{2}+\mathrm{x}+\frac{1}{4}-\frac{1}{4}+\mathrm{y}^{2}=0$
$\left(\mathrm{x}+\frac{1}{2}\right)^{2}+\mathrm{y}^{2}=\left(\frac{1}{2}\right)^{2}$
$\mathrm{x}^{2}+\mathrm{x}+\frac{1}{4}-\frac{1}{4}+\mathrm{y}^{2}=0$
$\left(\mathrm{x}+\frac{1}{2}\right)^{2}+\mathrm{y}^{2}=\left(\frac{1}{2}\right)^{2}$
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.