The circle passing through (1,-2) and touching the axis of x at (3,0) also passes through the point:

1) (5,-2)

2) (-2,5)

3) (-5,2)

4) (2,-5)

Question

The circle passing through (1,-2) and touching the axis of x at
(3,0) also passes through the point:
1) (5,-2)
2) (-2,5)
3) (-5,2)
4) (2,-5)

Ayush Jha · Accepted Answer

The equation of x axis

L: y=0

Now considering the equation of point circle as

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

So the equation of the family of the circle touching a given straight line(L: y=0) at a fixed point (3,0) is:

${(x - 3)}^{2} + {(y - 0)}^{2} + λ (y) = 0$

This circle passes through (1,-2). So,

${(1 - 3)}^{2} + {(- 2 - 0)}^{2} - 2 λ = 0 λ = 4$

So the equation of circle is-

${(x - 3)}^{2} + {(y)}^{2} + 4 (y) = 0 x^{2} + y^{2} + 4 y - 6 x + 9 = 0$

So, (5,-2) is the only point which satisfy the equation.

So the answer is (1) 5,-2