Question: Find the equation of a circle whose centre is (6,-3) and touches the Y-axis.

Solution:

In a circle,
Coordinates of the centre (h,k) = (6,-3)
Passing point lies on Y-axis.

We know,
When a circle touches Y-axis, its radius (r) = |h| units.
or, r = |h| units
or, r = |6|
So, r = 6 units

Now,
Using formula for circle equation, we get,
(x - h)² + (y - k)² = r²

or, (x -6)² + {y - (-3)}² = 6²

or, (x² - 12x + 36) + (y +3)² = 36

or, (x² - 12x) + (y² + 6y + 9) = 0

or, x² + y² - 12x + 6y + 9 = 0

Hence, the required equation of the given circle is x² + y² - 12x + 6y + 9 = 0.

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