Question

Equations of circles which pass through the points $(1,-2)$ and $(3,-4)$ and touch the $x-$axis is

• A. $x^2+y^2+6x+2y+9=0$
• B. $x^2+y^2+10x+20y+25=0$
• C. $x^2+y^2-6x+4y+9=0$
• D. $none$

Answer 1

Answer: (B), (C)

$x^2+y^2+10x+20y+25=0$
$x^2+y^2-6x+4y+9=0$

The general equation of a circle: $x^2+y^2+2gx+2fy+c=0$ or S=0if you have to find the equation of a circle when you have coordinates of two points through

which circle is passing then $S+tL=0$ is the required equation of your circle.

t is a parameter.

here S is the equation of the circle with given two points as diameter and

L is the equation of the line joining these two points.

The equation of the line passing through (1,-2) and (3,-4) is

L: $x+y+1=0$

The equation of the circle with (1,-2) and (3,-4) as diametrically opposite points is

S: $(x-1)(x-3)+(y+2)(y+4)=0$

S=$x^2+y^2-4x+6y+11=0$

Hence required equation of circle is$S^1=$ $x^2+y^2-4x+6y+11+t(x+y+1)=0$

Now to find t, we have the condition that circle also touches x-axis.

If a circle touches X-axis then $g^2=c$ for that circle.

For $S^1$,$g=\frac{t-4}{2}$and $c=t+11$

hence${\frac{(t-4)}{2}}^2=t+11$

$t^2-12t-28=0$

$(t-14)(t+2)=0$

$t=14,-2$

required equations are:$x^2+y^2+10x+20y+25=0$ and $x^2+y^2-6x+4y+9=0$