Question

The equation of circle which passes through $(1,-1)$ and which touches the line $6x+y-18=0$ at point $(3,0)$ is

• A. $13(x^2+y^2)-48x+5y+27=0$
• B. $13(x^2+y^2)-48x-5y+27=0$
• C. $11(x^2+y^2)-48x+5y+27=0$
• D. $13(x^2+y^2)+48x+5y+27=0$

Answer 1

The correct option is A $13(x^2+y^2)-48x+5y+27=0$

Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$ ------ (i)

since circle passes through $(1,-1)$

$1+1+2g-2f+c=0$

$g-f+(\frac{c}{2}+1)=0$ ---- (ii)

Also circle passes through $(3,0)$

$9+0+6g+c=0$

$g=-\frac{(c+9)}{6}$

Equation of tangent to the circle at point $(3,0)$ is

$x\times 3+y\times 0+g(x+3)+f(y+0)+c=0$

$(3+g)x+fy+3g+c=0$

but equation of tangent is given as $6x+y-18=0$

Therefore, $\frac{3+g}{6}=\frac{f}{1}=-\frac{18}{3g+c}$

$\frac{3+g}{6}=f$ and $f=-\frac{18}{3g+c}$

$f=\frac{3+g}{6}=\frac{1}{6}[3-\frac{(c+9)}{6}]$

$=\frac{1}{36}[9-c]$

putting the values of $f$ and $g$ in (ii)

$-\frac{(c+9)}{6}-\frac{(9-c)}{36}+\frac{c}{2}+1=0$

$c=\frac{27}{13}$

therefore, $g=-\frac{(c+9)}{6}=-24$

$f=\frac{1}{36}(9-c)=\frac{5}{26}$

putting the values of $f,g,c$ in equation (i),

$x^2+y^2+2gx+2fy+c=0$

We get,

$13(x^2+y^2)-48x+5y+27=0$

Hence the required equation of circle,

$13(x^2+y^2)-48x+5y+27=0$