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

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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$
- 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.

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