Question

The equation of the circle, having the lines $x^2+2xy+3x+6y=0$ as its normals and having size just sufficient to contain the circle $x(x-4)+y(y-3)=0$, is

• A. $x^2+y^2+6x+3y-45=0$
• B. $x^2+y^2+6x-3y-45=0$
• C. $x^2+y^2+6x-3y+45=0$
• D. None of these

Answer 1

The correct option is B $x^2+y^2+6x-3y-45=0$

Given equation of line is

$x^2+2xy+3x+6y=0$ ...(1)

$\Rightarrow x(x+2y)+3(x+2y)=0\Rightarrow (x+2y)(x+3)=0$

So, equations of normals are $x+3=0$ ...(2)

and $x+2y=0$ ...(3)

Solving (2) and (3), we get $x=-3,y=\frac{3}{2}$

$\therefore$Coordinates of centre of the circle are $(-3,\frac{3}{2}).$

Given equation of circle is $x(x-4)+y(y-3)=0$

$\Rightarrow x^2+y^2-4x-3y=0.$ ...(4)

Radius of the circle $=\sqrt{{(-2)}^2+{\left(\frac{-3}{2}\right)}^2}=\frac{5}{2}$ and coordinates of center $(2,\frac{3}{2}).$

Since the required circle is just sufficient to contain the circle (4), therefore the distance between the centres $(-3,\frac{3}{2})$ and $(2,\frac{3}{2})=$ the difference of their radius.

Let the radius of the required circle be $a.$

$\therefore \sqrt{{(-3-2)}^2+{(\frac{3}{2}-\frac{3}{2})}^2}=a-\frac{5}{2}\Rightarrow 5=a-\frac{5}{2}\Rightarrow a=5+\frac{5}{2}=\frac{15}{2}.$

Therefore, equation of required circle is

${(x+3)}^2+{(y-\frac{3}{2})}^2={\left(\frac{15}{2}\right)}^2$

$\Rightarrow x^2+y^2+6x-3y+9+\frac{9}{4}-\frac{225}{4}=0$

$\Rightarrow x^2+y^2+6x-3y-45=0.$