Find the equation of the circle with centre (–a, –b) and radius $$\sqrt{a^2-b^2}$$.

Asked by Abhisek | 1 year ago |  169

##### Solution :-

Centre (-a, -b) and radius $$\sqrt{a^2-b^2}$$

Let us consider the equation of a circle with centre (h, k) and

Radius r is given as (x – h)+ (y – k)= r2

So, centre (h, k) = (-a, -b) and radius (r) = $$\sqrt{a^2-b^2}$$

The equation of the circle is

(x + a)2 + (y + b)2 = $$(\sqrt{a^2-b^2})^2$$

x2 + 2ax + a2 + y2 + 2by + b2 = a2 – b2

x2 + y2 +2ax + 2by + 2b2 = 0

The equation of the circle is x2 + y2 +2ax + 2by + 2b2 = 0

Answered by Pragya Singh | 1 year ago

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