Determine the equation of the circle with radius 4 and Centre (-2, 3).

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Given that:

Radius, r = 4, and center (h, k) = (-2, 3).

We know that the equation of a circle with centre (h, k) and radius r is given as

(x – h)^2+ (y – k)^2= r^2….(1)

Now, substitute the radius and center values in (1), we get

Therefore, the equation of the circle is

(x + 2)^2+ (y – 3)^2= (4)^2

x^2+ 4x + 4 + y^2– 6y + 9 = 16

Now, simplify the above equation, we get:

x^2+ y^2+ 4x – 6y – 3 = 0

Thus, the equation of a circle with center (-2, 3) and radius 4 is x^2+ y^2+ 4x – 6y – 3 = 0

Thanks