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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