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

Asked by Abhisek | 11 months ago |  68

##### Solution :-

The centre of the circle is given as (h, k) = (2,2)

We know that the circle passes through point (4,5), the radius (r) of the circle is the distance between the points (2,2) and (4,5).

r = $$\sqrt{(2-4)^2 + (2-5)^2}$$

$$\sqrt{(-2)^2 + (-3)^2}$$

$$\sqrt{4+9}$$

$$\sqrt{13}$$

The equation of the circle is given as

(x– h)2+ (y – k)2 = r2

(x –h)2 + (y – k)2 =$$( \sqrt{13})^2$$

(x –2)2 + (y – 2)2 = $$( \sqrt{13})^2$$

x2 – 4x + 4 + y– 4y + 4 = 13

x2 + y2 – 4x – 4y = 5

The equation of the required circle is x2 + y2 – 4x – 4y = 5

Answered by Pragya Singh | 11 months ago

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