The equation of circle touching the line 2x+3y+1=0 at the point (1,-1) and passing through the focus of the parabola y2=4x isa.3x2+3y2-8x+3y+5=0b.3x2+3y2+8x-3y+5=0c.x2+y2-3x+y+6=0d.none of these

Hi Menka,

This is based on the equation of family of circles.

Write the eqn of the family of the circles touching the line 2x+3y=-1 at the point (1,-1).

The eqn of which is (x-1)^"2 + (y+1)^2 + λ(2x+3y+1) = 0.

This eqn should pass through the focus (1,0) of the parabola y^2 = 4x.

So you will get lambda, and hence the eqn of the circle.

Best Regards,

Ashwin (IIT Madras).

The answer is option (a).                                                                                                                                                                                        Since the circle is passing through the point (1,-1) {as it is the point of contact of the line with the circle}, that point has to satisfy the equation of the circle. Substitute x=1 and y=-1 in all the options and you come to know that equation in option (a) gets satisfied. Thus, option (a) is the answer. Also, you can check your answer by substituting the coordinates of focus of the given parabola, which is (a,0) i.e. (1,0). You find that option (a) is the correct option.