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

Explanation

Let `(h, k)` be the centre of a circle with radius `a`.

Thus, its equation will be

`(x-h)^{2}+(y-k)^{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)`.

`therefore h=0, k=6`

`therefore` Equation of the circle `=(x-0)^{2}+(y-6)^{2}=a^{2}`

Also, `a=4`

Substituting the value of `a` in equation `(1):(x-0)^{2}+(y-6)^{2}=16`

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