Mathematics | VITEEE 2019 | Discussion Page

A circle has a radius of 3 and its center lies on the line y = x - 1. The equation of the circle, if it passes through (7, 3), is

$x^{2} +y^{2} +8x-6y+16=0$

$x^{2} +y^{2} -8x+6y+16=0$

$x^{2} +y^{2} -8x-6y-16=0$

$x^{2} +y^{2} -8x-6y+16=0$

Option D

Let the centre of the circle be (h, k). Since the centre lies on the line y: x - 1

.'. k = h-1 ..(1)

Since the circle passes through the point (7, 3), therefore, the distance of the centre from this point is the radius of the circle.

.'. 3 = $\sqrt{(h-7)^{2}+(k-3)^{2}}$

3 = $\sqrt{(h-7)^{2}+(h-1-3)^{2}}$ using (1)

h = 7 or h = 4

F or h = 7, we get k = 6 and for h = 4, we get k = 3 Hence, the circles which satisfy the given conditions are:

$(x-7)^{2} +(y-6)^{2} =9$ or

$x^{2} + y^{2} -14x +12y +76 = 0$

and $(x -4)^{2} + (y- 3)^{2} = 9$ or

$x^{2} +y^{2} -8x-6y+16=0$

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