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Equation of Family of Circles Touching a Line and Passing through a Given Point on the Line

A circle pass...

Question

A circle passes through the point $(3, \sqrt{\frac{7}{2}})$ and touches the line pair $x^{2} - y^{2} - 2 x + 1 = 0$ . The coordinates of the centre of the circle are

(0, 4)

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(5, 0)

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(6, 0)

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none of these

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Solution

The correct option is A
(0, 4)

$(x - 1)^{2} - y^{2} = 0$

x + y - 1 = 0

x - y - 1 = 0

The bisector which bisects the angle containing the point $(3, \sqrt{\frac{7}{2}})$ is, y = 0 centre = (h, 0)

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

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A circle passes through the point (3,√72) and touches the line pair x2−y2−2x+1=0. The coordinates of the centre of the circle are

The correct option is A (0, 4) (x−1)2−y2=0 x + y - 1 = 0 x - y - 1 = 0 The bisector which bisects the angle containing the point (3,√72) is, y = 0 centre = (h, 0) |h−1|√2=√(h−3)2+72

A circle passes through the point $(3, \sqrt{\frac{7}{2}})$ and touches the line pair $x^{2} - y^{2} - 2 x + 1 = 0$ . The coordinates of the centre of the circle are

The correct option is A
(0, 4)

$(x - 1)^{2} - y^{2} = 0$

x + y - 1 = 0

x - y - 1 = 0

The bisector which bisects the angle containing the point $(3, \sqrt{\frac{7}{2}})$ is, y = 0 centre = (h, 0)

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