Question

Find the equation of a circle which of center is (2, 2) and which passes through the point (4,5)​

Answer 1

$\large\sf\underline{Correct\:Question}$

Find the equation of circle with centre (2,2) and passes through the point (4,5).

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‎$\large\sf\underline{Answer}$

${\sf{{\red{➻\:x^{2}+y^{2}-4x-4y-5=0}}}}$

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$\large\sf\underline{Solution}$

The centre of the circle is given as :

• ( h , k ) = ( 2 , 2 )

Since the circle passes through the point ( 4 , 5 ) , the radius ( r ) of the circle is the distance between the point ( 2 , 2 ) and ( 4 , 5 ) .

⠀⠀⠀

So we know ,

$\small{\underline{\boxed{\mathrm\pink{➻\:Radius=\sqrt{(a-h)^{2}+(b-k)^{2}}}}}}$

Here , a = 2

⠀⠀⠀⠀b = 2

⠀⠀⠀⠀h = 4

⠀⠀⠀⠀k = 5

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Now substituting the values we get radius as :

$\sf➞\:r=\sqrt{(2-4)^{2}+(2-5)^{2}}$

$\sf➞\:r=\sqrt{(-2)^{2}+(-3)^{2}}$

$\sf➞\:r=\sqrt{4+9}$

${\sf{{\red{➞\:r=\sqrt{13}}}}}$

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Thus the equation of the circle is :

$\small{\underline{\boxed{\mathrm\pink{(x-h)^{2}+(y-k)^{2}=r^{2}}}}}$

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Substituting the values we get the equation as :

$\sf➞\:(x-2)^{2}+(y-2)^{2}=[\sqrt{13}]^{2}$

$\sf➞\:x^{2}-4x+4+(y^{2}-4y+4)=13$

$\sf➞\:x^{2}-4x+4+y^{2}-4y+4=13$

$\sf➞\:x^{2}+y^{2}-4x-4y+4+4-13=0$

$\sf➞\:x^{2}+y^{2}-4x-4y+8-13=0$

$\small{\underline{\boxed{\mathrm\red{➻\:x^{2}+y^{2}-4x-4y-5=0}}}}$

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!! Hope it helps !!