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

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

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Let the equation of a circle be:

(x-h)^2+(y-k)^2=r^2

Now, since the centre of the circle is (2, 2), our equation becomes

(x-2)^2+(y-2)^2=r^2

Now, since this passes through the point (4, 5)

(4-2)^2+(5-2)^2=r^2

4+9=r^2

r^2=13

Hence, the final equation of the circle becomes

(x-2)^2+(y-2)^2=13

x^2-4x+4+y^2-4y+4=13

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

Posted by

Pankaj Sanodiya

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