Question: Find the equation of circle whose radius is √58 units and the centre lies on the x-axis and passing through the point (5,7).

Solution:
Given,
In a circle,
radius (r) = √58 units
passing point (x2,y2) = (5,7)

We know,
Radius is the distance between the centre of the circle and the passing point.

Let the centre of the circle be denoted by (x1,y1). Since, the centre lies on x-axis, the coordinates become (a,0). So, (x1,y1) = (a,0) using formula,

$r = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

$or, r^2 = (5 - a)^2 + (7 - 0)^2}$

$or, (\sqrt{58})^2 = 25 - 10a + a^2 + 49$

$or, 58 = 74 - 10a + a^2$

$or, a^2 - 10a + 16 = 0$

$or, a^2 - 2a - 8a +16 = 0$

$or, a(a -2) - 8(a -2) = 0$

$or, (a -8)(a -2) = 0$

Either, a = 8 Or, a = 2

When centre (h,k) = (8,0) and r = √58 units, equation of circle is
(x - h)² + (y - k)² = r²
or, (x - 8)² + (y -0)² = 58
or, (x² - 16x + 64) + y² = 58
or, x² + y² - 16x + 6 = 0 is the required equation.

When centre (h,k) = (2,0) and r = √58 units, equation of circle is
(x - h)² + (y - k)² = r²
or, (x - 2)² + (y -0)² = 58
or, x² -4x + 4 + y² = 58
or, x² + y² - 4x - 54 = 0 is the required equation.

Hence, the required equation of the given circle is either (x² + y² - 16x + 6 = 0) or (x² + y² - 4x - 54 = 0).

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