Question

The general form of the equation of a circle is 7x2 + 7y2 − 28x + 42y − 35 = 0. The equation of this circle in standard form is . The center of the circle is at the point , and its radius is units.

Answer 1

Answer:

The standard form of a circle is (x-h)^2 + (y-k)^2 = r^2 with (h,k) being the center of the circle and r being the radius. In this case the circle's equation in standard form is (x-2)^2 + (y+3)^2 = 18. Knowing this it's easy to see that the center of the circle (h,k) is (2,-3). Finally the radius is $\sqrt{18}$ or in simplified terms, 3$\sqrt{2}$

Step-by-step explanation:

Answer 2

Answer:

see explanation

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

Given

7x² + 7y² - 28x + 42y - 35 = 0 ( divide through by 7 )

x² + y² - 4x + 6y - 5 = 0

Collect x- terms and y- terms together and add 5 to both sides

x² - 4x + y² + 6y = 5

Use the method of completing the square to obtain standard form

add ( half the coefficient of the x / y term)² to both sides

x² + 2(- 2)x + 4 + y² + 2(3)y + 9 = 5 + 4 + 9

(x - 2)² + (y + 3)² = 18 ← in standard form

here (h, k) = (- 2, 3) and r = $\sqrt{18}$ = 3$\sqrt{2}$

Centre = (2, - 3) and radius = 3$\sqrt{2}$