Question

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Answer 1

Given that:

Radius, r = 8, and center (h, k) = (2, -3).

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

\sf (x – h)^2 + (y – k)^2 = r^2 ….(1)(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

sf (x + 2)^2+ (y – 3)^2 = (8)^2(x+2)

2 +(y–3) 2 =(8) 2

\sf x^2+ 4x + 4 + y^2 – 6y + 9 = 64x 2 +4x+4+y 2 –6y+9=64

Now, simplify the above equation, we get:

sf x^2 + y^2+ 4x – 6y – 53= 0 x 2 +y 2 +4x–6y–53=0

Thus, the equation of a circle with center (-2, 3) and radius 8 is \sf x^2 + y^2+ 4x –6y– 53 = 0 x 2 +y 2+4x–6y–53=0

.....hope it helps uhh mate....

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