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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the equation of the circle with radius $5$ whose centre lies on x-axis and passes through the point $(2,3)$.

$\begin {array} {1 1} (A)\;x^2+y^2+4x-21=0 \: and \: x^2+y^2-12x+11=0 \\ (B)\;x^2+y^2-4x+21=0 \: and \: x^2+y^2+12x-11=0 \\ (C)\;x^2+y^2-4x-21=0 \: and \: x^2+y^2-12x-11=0 \\ (D)\;x^2+y^2+4x+21=0 \: and \: x^2+y^2+12x+11=0 \end {array}$

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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 $(x-h)^2+(y-k)^2=r^2$
It is given that the radius of the circle is 5 and its centre lies on the x - axis , (i.e) k = 0 and r = 5.
Hence the equation of the circle is $ ( x-h)^2+y^2=25$
It is also given that the circle passes through the point (2,3)
$ \therefore (2-h)^2+3^2=25$
$(2-h)^2=16$
taking square root on both sides we get,
$ 2 - h = \pm 4$
Hence $ h = -2 $ and $6$
Case (i)
When $h= -2$, then the equation of the circle is
$(x+2)^2+y^2=25$
On expanding we get,
$x^2+4x+4+y^2=25$
(i.e) $x^2+y^2+4x-21=0$ is the required equation of the circle.
Case (ii)
When h = 6
$(x-6)^2+y^2=25$
On expanding we get,
$x^2-12x+36+y^2=25$
(i.e) $x^2+y^2-12x+11=0$ is the required equation of the circle.
answered Jul 13, 2014 by thanvigandhi_1
 

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