Question: Find the equation of the circle which touches the X-axis at a point (3,0) and passing through the point (1,2).

Solution:

In a circle,
It touches the X-axis at a point (3,0).
Coordinates of the centre (h,k) = (3,k)
So, radius (r) = |k| units
Passing point (x,y) = (1,2)

Now,
radius = distance between centre and passing point
$or, r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
$or, k = \sqrt{(1 - 3)^2 + (2 - k)^2}$
$or, k = \sqrt{4 + 4 - 4k + k^2}$
[Squaring both sides]
$or, k^2 = ( \sqrt{8 - 4k + k^2})^2$
$or, k^2 = k^2 - 4k + 8$
$or, 4k = 8$
$\therefore k = 2$

So, centre (h,k) = (3,2) and radius = 2 units

Now, equation of circle is given by;
(x - h)² + (y - k)² = r²
or, (x - 3)² + (y - 2)² = 2²
or, (x² - 6x + 9) + (y² - 4y +4) = 4
or, x² + y² - 6x - 4y +9 = 0 is the required equation.

Hence, the required equation of the given circle is x² + y² - 6x - 4y + 9 = 0.

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