The equation of the circle with centre $(2, 2)$ which passes through $(4, 5)$ is

x^{2} + y^{2} - 4 x + 4 y - 77 = 0

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x^{2} + y^{2} - 4 x - 4 y - 5 = 0

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x^{2} + y^{2} + 2 x + 2 y - 59 = 0

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x^{2} + y^{2} - 2 x - 2 y - 23 = 0

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x^{2} + y^{2} + 4 x - 2 y - 26 = 0

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Solution

The correct option is E $x^{2} + y^{2} - 4 x - 4 y - 5 = 0$
Radius of circle is $\sqrt{(4 - 2)^{2} + (5 - 2)^{2}} = \sqrt{13}$ So, equation of circle is
$(x - 2)^{2} + (y - 2)^{2} = 13$
$\Rightarrow x^{2} + 4 - 4 x + y^{2} + 4 - 4 y = 13$
$\Rightarrow x^{2} + y^{2} - 4 x - 4 y - 5 = 0$

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The equation of the circle with centre (2,2) which passes through (4,5) is

The correct option is E x2+y2−4x−4y−5=0Radius of circle is √(4−2)2+(5−2)2=√13 So, equation of circle is(x−2)2+(y−2)2=13⇒x2+4−4x+y2+4−4y=13⇒x2+y2−4x−4y−5=0

The equation of the circle with centre $(2, 2)$ which passes through $(4, 5)$ is

The correct option is E $x^{2} + y^{2} - 4 x - 4 y - 5 = 0$
Radius of circle is $\sqrt{(4 - 2)^{2} + (5 - 2)^{2}} = \sqrt{13}$ So, equation of circle is
$(x - 2)^{2} + (y - 2)^{2} = 13$
$\Rightarrow x^{2} + 4 - 4 x + y^{2} + 4 - 4 y = 13$
$\Rightarrow x^{2} + y^{2} - 4 x - 4 y - 5 = 0$