Question

Find the equation of the circle whose centre lies on the positive direction of y - axis at a distance 6 from the origin and whose radius is 4.

Answer 1

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be

\[\left( x - h \right)^2 + \left( y - k \right)^2 = a^2\]

The centre of the required circle lies on the positive direction of the y-axis at a distance 6 from the origin.
Thus, the coordinates of the centre are (0, 6).

∴ h = 0, k = 6
∴ Equation of the circle = \[\left( x - 0 \right)^2 + \left( y - 6 \right)^2 = a^2\]

Also, a = 4
Substituting the value of a in equation (1): \[\left( x - 0 \right)^2 + \left( y - 6 \right)^2 = 16\]

\[\Rightarrow x^2 + y^2 + 36 - 12y = 16\]
\[ \Rightarrow x^2 + y^2 - 12y + 20 = 0\]

Hence, the required equation of the circle is

\[x^2 + y^2 - 12y + 20 = 0\]