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

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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.

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