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 the circle passing through the points $(2,3)$ and $(–1,1)$ and whose centre is on the line $x–3y–11 = 0$.

$\begin {array} {1 1} (A)\;x^2+y^2+7x-5y+14=0 & \quad (B)\;x^2+y^2-7x-5y+14=0 \\ (C)\;x^2+y^2-7x-5y-14=0 & \quad (D)\;x^2+y^2-7x-5y+14=0 \end {array}$

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$
Step 1 :
Let the equation of the required circle be $ ( x-h)^2+(y-k)^2=r^2$-----(1)
It is given that the circle passes through the point (2,3) and (-1, 1)
$ \therefore (2-h)^2 +(3-k)^2=r^2$ and
$(-1-h)^2+(1-k)^2=r^2$
$ \Rightarrow (2-h)^2+(3-k)^2=(-1-h)^2+(1-k)^2$
On expanding
$4-4h+h^2+9-6k+k^2=1+2h+h^2+1-2k+k^2$
$ \Rightarrow 4-4h+9-6k=1+2h+1-2k$
$ \Rightarrow 6h+4k=11$------(2)
It is also given that the centre(h, k) of the circe lies on the line $x-3y-11=0$.
$ \therefore h - 3k=11$---------(3)
Step 2 :
Solving equation (2) and (3) we get,
$ h-3k=11 ( \times 6)$
$6h+4k=11$
_____________
$6h-18k=66$
$6h+4k=11$
$(-) \: \: (-) \: \: (-)$
_______________
$ \quad -22k=55$
$ \Rightarrow k = -\large\frac{5}{2}$
Substituting for k in equation (3) we get,
$ h = 3 \bigg( -\large\frac{5}{2} \bigg) $$=11$
$ \Rightarrow h = 11-\large\frac{15}{2}$
(i.e) $ h = \large\frac{7}{2}$
Hence the coordinates of the centre are $ \bigg( \large\frac{7}{2}$$, -\large\frac{5}{2} \bigg)$
Substituting for h and k in equation (1) we get,
$\bigg( 2 - \large\frac{7}{2} \bigg)^2+ \bigg( 3+ \large\frac{5}{2} \bigg)^2=r^2$
(i.e) $ \bigg( \large\frac{4-7}{2} \bigg)^2$$+ \bigg( \large\frac{6+5}{2} \bigg)^2$$=r^2$
(i.e) $ \large\frac{9}{4} $$+ \large\frac{121}{4}$$=r^2$
(i.e) $r^2=\large\frac{130}{4} \Rightarrow r = \large\frac{\sqrt{130}}{2}$
Hence the equation of the circle is
$ \bigg( x - \large\frac{7}{2} \bigg)^2$$+ \bigg( y+ \large\frac{5}{2} \bigg)^2$$=\large\frac{130}{4}$
On expanding we get,
$4x^2-28x+49+4y^2+20y+25=130$
$ \Rightarrow 4x^2+4y^2-28x+20y-56=0$
dividing throughout by 4 we get,
$x^2+y^2-7x+5y-14=0$
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
...