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. $x^2+y^2+3x-6y-40=0$
• B. $x^2+y^2+6x-3y-45=0$
• C. $x^2+y^2+8x+4y-20=0$
• D. $x^2+y^2+4x+8y+20=0$

Answer 1

Answer: (A)

$x^2+y^2+3x-6y-40=0$

Pair of lines:

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

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

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

$\Rightarrow x+2y=0$

and $x+3=0$

Are the two normals and their point of intersection must be the centre of the circle.

$\Rightarrow x=-3$

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

$\Rightarrow (-3,\frac{3}{2})$ is the centre of required circle.

($\because x(x-4)+y(y-3)=0$ is the diametric form of the circle. Hence, $(0,0)$ and $(4,3)$ are the diametric end points.)

$\Rightarrow$ Centre $\to (\frac{0+4}{2},\frac{0+3}{2})\to (2,\frac{3}{2})$

radius $=\frac{1}{2}\left(\sqrt{16+9}\right)$

$=\frac{5}{2}$

The required circle with centre $(-3,\frac{3}{2})$ is just sufficient to contain the circle

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

$\therefore$ radius of required circle

=distance between $(-3,\frac{3}{2})$ and $(2,\frac{3}{2})$ $+$ radius of given circle.

$=\sqrt{(-3-2{)}^2+0}+\frac{5}{2}$

$=5+\frac{5}{2}=\frac{15}{2}$

$\therefore$ Equation of required circle:

$\Rightarrow (x+3{)}^2+(y-\frac{3}{2}{)}^2=\frac{225}{4}$

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

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

$\Rightarrow x^2+y^2+6x-3y+9-\frac{216}{4}=0$

$\Rightarrow x^2+y^2+6x-3y+9-54=0$

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

$\therefore B)$ Answer.