Correct Answer - Option 3 : 5 and (0, -3)
Concept:
The general equation of a second-degree curve is
\(\rm ax^2+2hxy+by^2+2gx+2fy+c =0\)
Where a, b, c, f, g, h are constant
If a = b ≠ 0 and h = 0, then it represents a circle.
The general equation for the circle
\(\rm ax^2+ay^2+2gx+2fy+c =0\)
The radius of that circle = \(\rm \sqrt{\left({g\over a}\right)^2+\left({f\over a}\right)^2-\left({c\over a}\right)}\)and center \(\rm \left(-{g\over a},-{f\over a}\right)\)
Calculation:
Given equation 2y2 + 2x2 + 12y = 32
⇒ 2y2 + 2x2 + 12y - 32 = 0
Comparing to the general equation of a circle
\(\rm ax^2+ay^2+2gx+2fy+c =0\)
a = 2, g = 0, f = 6, c = - 32
∴ Center = \(\rm \left(-{g\over a},-{f\over a}\right)\) = (0, -3)
and Radius = \(\rm \sqrt{\left({g\over a}\right)^2+\left({f\over a}\right)^2-\left({c\over a}\right)}\)
⇒ R = \(\rm \sqrt{\left({0\over 2}\right)^2+\left({6\over 2}\right)^2-\left({-32\over 2}\right)}\)
⇒ R = \(\rm \sqrt{\left({0\over 2}\right)^2+\left({6\over 2}\right)^2+\left({32\over 2}\right)}\)
⇒ R = \(\rm \sqrt{\left(9+16\right)}\) = 5
