Find the centre and radius of the circle x2 + y2 – 8x + 10y – 12 = 0

Asked by Abhisek | 1 year ago |  125

##### Solution :-

The equation of the given circle is x2 + y2 -8x + 10y -12 = 0.

x2 + y2 – 8x + 10y – 12 = 0

(x2 – 8x) + (y+ 10y) = 12

(x2 – 2(x) (4) + 42) + (y2 – 2(y) (5) + 52) – 16 – 25 = 12

(x – 4)2 + (y + 5)2 = 53

(x – 4)2 + (y – (-5))2 = $$( \sqrt{53})^2$$ [which is form (x-h)2 +(y-k)2 = r2]

Where h = 4, K= -5 and r = $$\sqrt{53}$$

The centre of the given circle is (4, -5) and its radius is $$\sqrt{53}$$.

Answered by Pragya Singh | 1 year ago

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