Equation of the circle which is such that the length of the tangents to it from the points (1,0), (0, 2) and (3, 2) are $1, \sqrt{7}$ respectively is

6 (x^{2} + y^{2}) - 28 x - 5 y + 28 = 0

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9 (x^{2} + y^{2}) - 28 x - 5 y + 28 = 0

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3 (x^{2} + y^{2}) - 28 x - 5 y + 28 = 0

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x^{2} + y^{2} - 28 x - 5 y + 28 = 0

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Solution

The correct option is A $6 (x^{2} + y^{2}) - 28 x - 5 y + 28 = 0$

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Similar questions

Find the equation of the circle orthogonal to the circles $x^{2} + y^{2} + 3 x - 5 y + 6 = 0 a n d 4 x^{2} + 4 y 2 - 28 x + 29 = 0$ and whose center lies on the line 3x + 4y + 1 = 0.

Q. Obtain the equation of the circle orthogonal to both the circles

x^{2} + y^{2} + 3 x - 5 y + 6 = 0

and

4 x^{2} + 4 y^{2} - 28 x + 29 = 0

and whose centre lies on the line

3 x + 4 y + 1 = 0

Q. P is a variable point and the co-ordinates of two points A and B are (–2, 2, 3) and (13, –3, 13) respectively. If 3PA = 2PB, then locus of P is :

Find the equation of the circle orthogonal to the circles $x^{2} + y^{2} + 3 x - 5 y + 6 = 0 a n d 4 x^{2} + 4 y 2 - 28 x + 29 = 0$ and whose center lies on the line 3x + 4y + 1 = 0.

Q. Equation of the circle having centre at

(3, - 1)

and making an intercept of length

6

units on the line

2 x - 5 y + 18 = 0

, is

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Equation of the circle which is such that the length of the tangents to it from the points (1,0), (0, 2) and (3, 2) are 1,√7 respectively is

The correct option is A 6(x2+y2)−28x−5y+28=0

Equation of the circle which is such that the length of the tangents to it from the points (1,0), (0, 2) and (3, 2) are $1, \sqrt{7}$ respectively is

The correct option is A $6 (x^{2} + y^{2}) - 28 x - 5 y + 28 = 0$