Find the centre and radius of the circle x2 + y2 – 4x – 8y – 45 = 0

Asked by Abhisek | 1 year ago |  109

#### 1 Answer

##### Solution :-

The equation of the given circle is x2 + y2 – 4x – 8y – 45 = 0.

x2 + y2 – 4x – 8y – 45 = 0

(x2 – 4x) + (y2 -8y) = 45

(x2 – 2(x) (2) + 22) + (y2 – 2(y) (4) + 42) – 4 – 16 = 45

(x – 2)2 + (y – 4)2 = 65

(x – 2)2 + (y – 4)2 = $$( \sqrt{65})^2$$ [which is form (x-h)2 +(y-k)2 = r2]

Where h = 2, K = 4 and r =$$\sqrt{65}$$

The centre of the given circle is (2, 4) and its radius is $$\sqrt{65}$$.

Answered by Pragya Singh | 1 year ago

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