Question

The equation of the circle passing through the
Point (-1,-3) and touching the line
4x+3y - 12 = 0 at the point (3,0) is.
a) x? +y? - 2x +3y +3 = 0
b) x² + y2 - 2x + 3y - 3 = 0
c) x + y² +2x+3y+3 = 0
d) x² + y2 +x+y+1=0​

Answer 1

Given:

The circle passes through point (-1,-3) and touches the line 4x+3y-12 = 0 at the point (3,0)

To find:

We have to find the equation of circle

Solution:

Answer:  (b) $x^{2}+y^{2}-2x+3y-3 =0$

The general equation of circle is given by  $x^{2} +y^{2}+2gx+2fy+c=0$

And centre of circle is $(-g,-f)$

As it is given that circle passes through (-1,-3) and (3,0) substitute these points in general equation of circle

we get,$(-1)^{2}+(-3)^{2}+2g(-1) +2f(-3)+c=0$

            $1+9-2g-6f+c=0$

             $2g+6f-c=10$ _ _(1)

And  $3^{2}+0^{2}+2g(3)+2f(0)+c=0$

        $6g+c=-9$ _ _(2)

Now,(1)+(2)⇒ $8g+6f=1$ _ _(3)

As the given line just touches the circle it acts as a tangent to it.

we know that any tangent is perpendicular to the radius through the point of contact.

So now radius through the centre  $(-g,-f)$ and point of contact (3,0)   as well as the line 4x+3y - 12 = 0 are perpendicular

Now considering the slope of line as m₁ and considering the slope of radius as m₂ we can write m₁m₂ = -1

where $m_1=\frac{-4}{3}$  and   $m_2=\frac{f}{3+g}$

           $(\frac{-4}{3})(\frac{f}{3+g})=-1$

            $4f=9+3g$

           $3g-4f=-9$ _ _(4)

Solving (3) and (4) we get  $f=\frac{3}{2}$  and  $g= -1$

Substitute  g and f values in (1)

$2(-1) +6(\frac{3}{2} )-c=10$

$7-c=10$

$c=-3$

Now substitute $g,f,c$ values in general equation of circle

$x^{2}+y^{2}+2(-1)x+2(\frac{3}{2} )y+(-3) =0$

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

∴ The equation of the required circle is $x^{2}+y^{2}-2x+3y-3 =0$