Question

Find the equation of the circle touching the line $2x+3y+1$ = 0 at the point $(1,-1)$ and is orthogonal to the circle which has the line segment having end points($0$, $-1$) and $(-2,3)$ as the diameter.

• A. $2x^2+4y^2+5x-5y+1=0$
  
• B. $2x^2+3y^2-10x+5y+4=0$
• C. $3x^2+2y^2+5x-5y+2=0$
• D. $2x^2+2y^2-10x-5y+1=0$

Answer 1

The correct option is D $2x^2+2y^2-10x-5y+1=0$

Circle having $(0,1)$ and $(-2,3)$ as diameter:

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

$\Rightarrow x^2+2x+y^2-2y-3=0$

$\Rightarrow x^2+y^2+2x-2y-3=0\to \left(1\right)$

Let the required circle be

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

This circle is orthogonal to $\left(1\right)$

$\Rightarrow 2g\left(1\right)+2f(-1)=c-3$

$\Rightarrow 2g-2f=c-3\to \left(2\right)$

Tangent to $x^2+y^2+2gx+2fy+c=0$ at $(1,-1)$ is $T=0$

$\Rightarrow x\left(1\right)+y(-1)+g(x+1)+y(f-1)+c=0$

$\Rightarrow x-y+gx+g+fy-y+c=0$

$\Rightarrow x(g+1)+y(f-2)+(g+c)=0$

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

This line equation is

$2x+3y+1=0$

$\Rightarrow g+1g+c=2$

$\Rightarrow g+1=2g+2c$

$\Rightarrow g=1-2c\to \left(3\right)$

and $\frac{f-2}{g+c}=3$

$\Rightarrow f-2=3g+3c\to \left(4\right)$

Solving $\left(2\right)$ and $\left(3\right)$

$\Rightarrow 2(1-2c)-2f=c-3$

$\Rightarrow 2-4c-2f=c-3$

$\Rightarrow 5c-5=-2f$

$\Rightarrow 5c+2f=5\to \left(5\right)$

Solving $\left(3\right)$ and $\left(4\right)$

$\Rightarrow f-2=3(1-2c)+3c$

$\Rightarrow f-2=3-6c+3c$

$\Rightarrow f-2=3-3c$

$\Rightarrow 3c+f=5\to \left(6\right)$

Solving $\left(5\right)$ and $\left(6\right)$

$\Rightarrow 5c+2(5-3c)=5$

$\Rightarrow 5c+10-6c=5$

$\Rightarrow 5=c$

$\Rightarrow c=5$

Hence , $f=5-3c=-10$

$g=1-2c=1-10=-9$

$\therefore$ Circle $\to x^2+y^2-18x-20y+5=0$

$\therefore$ None of these options.