# Find the equation of circle having the lines $$x^2$$ + 2xy + 3x + 6y = 0 as its normal and having size just sufficient to contain the circle x(x – 4) + y(y – 3) = 0.

## Solution :

Pair of normals are (x + 2y)(x + 3) = 0

$$\therefore$$ Normals are x + 2y = 0, x + 3 = 0

Point of intersection of normals is the center of the required circle i.e. $$C_1$$(-3,3/2) and center of the given circle is

$$C_2$$(2,3/2) and radius $$r^2$$ = $$\sqrt{4 + {9\over 4}}$$ = $$5\over 2$$

Let $$r_1$$ be the radius of the required circle

$$\implies$$  $$r_1$$ = $$C_1$$$$C_2$$ + $$r_2$$ = $$\sqrt{(-3-2)^2 + ({3\over 2}- {3\over 2})^2}$$ + $$5\over 2$$ = $$15\over 2$$

Hence equation of required circle is $$x^2 + y^2 + 6x – 3y – 45$$ = 0

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