Find x intercepts of (x-4)^2+(y-3)^2=81?

Question

center will be 8, =3 radius = 9 
can you tell me if the information is complete?

Answer 1

The equation is (x - 4)^2 + (y - 3)^2 = 81.

To find x - intercepts, substitute y = 0 in the given equation.

(x - 4)^2 + ( 0 - 3)^2 = 81

(x - 4)^2 + 9 = 81

(x - 4)^2 = 81 - 9

(x - 4)^2 = 72

x^2 - 8x + 16 - 72 = 0

x^2 - 8x - 56 = 0.

x^2 - 8x - 56 = 0, is a quadratic, so use quadratic formula to find the roots of related quadratic equation.

Compare the above equation with standard form of the quadratic equation ax^2 + bx + c = 0.

a = 1, b = - 8, and c = - 56.

The solution x = [ - b ± √(b^2 - 4ac) ] / 2a.

Substitute the values b = - 8, a = 1, and c = - 56.

x = [ - (- 8) ± √((- 8)^2 - 4(1)(- 56)) ] / 2(1)

x = [ 8 ± √(64 + 224) ] / 2

x = [ 8 ± √288] / 2

x = 4 ± 6√2.

Therefore, x - intercepts of the given equation are x = 4 + 6√2 and x = 4 - 6√2.