Question

Select the correct answer from each drop-down menu. 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

The center of the circle is at the point (2, -3), and its radius is 4 units

Explanation:

The equation of the circle in standard form can be obtained by completing the square for both x and y terms as follows:

7x^2 - 28x + 7y^2 + 42y - 35 = 0

7(x^2 - 4x) + 7(y^2 + 6y) = 35

7(x^2 - 4x + 4 - 4) + 7(y^2 + 6y + 9 - 9) = 35

7[(x - 2)^2 - 4] + 7[(y + 3)^2 - 9] = 35

7(x - 2)^2 + 7(y + 3)^2 - 112 = 0

7(x - 2)^2 + 7(y + 3)^2 = 112

Dividing by 112 on both sides, we get:

(x - 2)^2 + (y + 3)^2 / 16 = 1

Therefore, the equation of the circle in standard form is (x - 2)^2 + (y + 3)^2 / 16 = 1.

The center of the circle is at the point (2, -3), and its radius is 4 units.

Answer 2

The general form of the equation of a circle is given by:

(x - h)^2 + (y - k)^2 = r^2

where (h, k) is the center of the circle and r is its radius.

Comparing the given equation to the standard form, we can complete the square for x and y to get:

7(x^2 - 4x) + 7(y^2 + 6y) = 35

7[(x - 2)^2 - 4] + 7[(y + 3)^2 - 9] = 35

7(x - 2)^2 + 7(y + 3)^2 = 98

Dividing both sides by 98, we get:

(x - 2)^2/14 + (y + 3)^2/14 = 1

Therefore, the equation of the circle in standard form is:

(x - 2)^2/14 + (y + 3)^2/14 = 1

The center of the circle is at the point (2, -3), and its radius is sqrt(14) units.