Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XI  >>  Math  >>  Conic Sections
0 votes

$x^2+y^2–8x+10y–12=0$, find the centre and radius of the circles.

$\begin {array} {1 1} (A)\;(4, -5) \: and \: \sqrt{53} \\ (B)\;(-4, -5) \: and \: \sqrt{53} \\ (C)\;(-4, 5) \: and \: \sqrt{53} \\ (D)\;(4, 5) \: and \: \sqrt{53} \end {array}$

Can you answer this question?

1 Answer

0 votes
  • Equation of a circle with centre (h,k) and radius r is given as $ (x-h)^2+(y-k)^2=r^2$
Equation of the given circle is
This can be rewritten as
(i.e) $[ (x-4)^2-16]+[(y+5)^2-25]=12$
(i.e) $(x-4)^2+(y+5)^2=53$
This is of the form
Comparing both the equations we get,
$h=4 , k =-5$ and $r = \sqrt{53}$
Hence the coordinates of the centre is
(4, -5) and $ \sqrt{53}$
answered Jul 13, 2014 by thanvigandhi_1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App