Question

The radius of circle with 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. $\frac{15}{2}$
• B. $\frac{13}{2}$
• C. $\frac{17}{2}$
• D. $5$

Answer 1

The correct option is A $\frac{15}{2}$

The combined equation of circles is given by,

$x^2+2xy+3x+6y=0$

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

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

$\therefore (x+2y)=0$ and $(x+3)=0$

$\therefore y=\frac{-1}{2}x$ and $x=-3$

Now, point of intersection of these two normals will be center of required circle.

Now, equation of one of the normals is given as $x=-3$

Put this value in second equation of normal, we get,

$-3+2y=0$

$\therefore 2y=3$

$\therefore y=\frac{3}{2}$

Thus, coordinates of center of required circle are $C_1(-3,\frac{3}{2})$

Now, equation of second circle is given as,

$x(x-4)+y(y-3)=0$

$\therefore x^2-4x+y^2-3y=0$

$\therefore x^2+y^2-4x-3y=0$

Compare this equation with standard equation of circle i.e. $\therefore x^2+y^2+2gx+2fy+c=0$, we get,

$2g=-4$

$\therefore g=-2$

$2f=-3$

$\therefore f=\frac{-3}{2}$

Thus, coordinates of this circle are $C_2(-g,-f)=C_2(2,\frac{3}{2})$

Radius of this circle is $r=\sqrt{g^2+f^2-c}$

$r=\sqrt{{(-2)}^2+{\left(\frac{-3}{2}\right)}^2-0}$

$r=\sqrt{4+\frac{9}{4}}$

$\therefore r=\sqrt{\frac{25}{4}}$

$\therefore r=\frac{5}{2}$

Let R = radius of required circle .

By the given condition, as required circle just contains circle $C_1$, we can use the following formula for circles just containing other circle-

$C_1{C}_2=R-r$ Equation (1)

Now, by distance formula,

$C_1{C}_2=\sqrt{{[2-(-3)]}^2+{(\frac{3}{2}-\frac{3}{2})}^2}$

$\therefore C_1{C}_2=\sqrt{{\left[5\right]}^2}$

$\therefore C_1{C}_2=5$

From equation (1), we can write,

$5=R-\frac{5}{2}$

$\therefore R=5+\frac{5}{2}$

$\therefore R=\frac{15}{2}$

Thus, answer is option (A)