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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x + y = 16

$\begin {array} {1 1} (A)\;x^2+y^2+6x+8y+15=0 & \quad (B)\;x^2+y^2-6x+8y+15=0 \\ (C)\;x^2+y^2-6x-8y-15=0 & \quad (D)\;x^2+y^2-6x-8y+15=0 \end {array}$

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  • 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 (4,1) and (6,5)
$ \therefore (4-h)^2+(1-k)^2=r^2$ and
$(6-h)^2+(5-k)^2=r^2$
$ \Rightarrow (4-h)^2+(1-k)^2=(6-h)^2+(5-k)^2$
On expanding we get,
$16-8h+h^2+1-2k+k^2=36-12h+h^2+25-10k+k^2$
$ \Rightarrow 16-8h+1-2k=36-12h+25-10k$
$ \Rightarrow 4h+8k=44$
Dividing throughout by 4 we get,
$h+2k=11$-------(2)
It is given that the centre of the circle lies on the line $4x+y=16$
$ \therefore 4h+k=16$-------(3)
Solving equation (2) and (3) for h and k we get,
$ \: h+2k=11 ( \times 4 )$
$ \not{4}\not{h}+k=16$
$\not{4}\not{h} + 8k = 44$
$(-) \quad (-) \quad (-)$
______________________
$ \qquad -7k = -25$
$ \Rightarrow k = 4$ Substituting for k in equation (1)
$h + 2(4)=11$
$h=3$
$ \therefore h =3$ and $ k = 4$
Substituting the values of h and k in equation (1), we get,
$(4-3)^2+(1-4)^2=r^2$
$1^2+(-3)^2=r^2$
$ \Rightarrow r^2=10$ or $r=\sqrt{10}$
Step 3 :
Hence the equation of the circle is
$(x-3)^2+(y-4)^2= ( \sqrt{10} )^2$
(i.e) $x^2-6x+9+y^2-8y+16=10$
(i.e) $x^2+y^2-6x-8y+15=0$
is the required equation of the circle.
answered Jul 13, 2014 by thanvigandhi_1
 

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