logo

Ask Questions, Get Answers

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

Find the equation of a circle with centre $(2,2)$ and passes through the point $(4,5)$.

$\begin {array} {1 1} (A)\;x^2+y^2-4x-4y-5=0 & \quad (B)\;x^2+y^2+4x-4y-5=0 \\ (C)\;x^2+y^2+4x+4y-5=0 & \quad (D)\;x^2+y^2+4x+4y+5=0 \end {array}$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Equation of a circle with centre (h,k) and radius r is given as $ (x-h)^2+(y-k)^2=r^2$
  • Distance between two points is $ \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Step 1 :
The coordinates of the centre are (2,2)
(i.e) (h,k)=(2,2)
Since it passes through (4,5) .
The radius of the circle is the distance between the points (2,2) and (4,5)
$ \therefore r = \sqrt{ 2-4)^2+(2-5)^2}$
$= \sqrt{4+9} = \sqrt{13}$
Hence the equation of the circle is
$(x-2)^2+(y-2)^2= ( \sqrt{13} )^2$
On expanding we get,
$x^2-4x+4+y^2-4y+4=13$
(i.e) $x^2+y^2-4x-4y-5=0$
This is the equation of the required circle.
answered Jul 13, 2014 by thanvigandhi_1
 

Related questions

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