Question

The equation of the circle having the lines ${x^2} + 2xy + 3x + 6y = 0$ as its normal having size just sufficient to contain the circle $x\left( {x - 4} \right) + y\left( {y - 3} \right) = 0$ is

1. ${x^2} + {y^2} + 3x - 6y - 40 = 0$
2. ${x^2} + {y^2} + 6x - 3y - 45 = 0$
3. ${x^2} + {y^2} + 8x + 4y - 20 = 0$
4. ${x^2} + {y^2} + 4x + 8y + 20 = 0$

Answer 1

Find the centre and the radius of the given circle $x\left( {x - 4} \right) + y\left( {y - 3} \right) = 0$. The radius of the required circle can be found by finding the intersection point of the normal lines ${x^2} + 2xy + 3x + 6y = 0$. The radius of the bigger circle can be found by using the equation
${C_1}{C_2} = R - r$.

Complete step-by-step answer:
It is known that the equation of the circle is given by the $\left( {x - {x_1}} \right)\left( {x - {x_2}} \right) + \left( {y - {y_1}} \right)\left( {y - {y_2}} \right) = 0$, where $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ are points on the extreme end of the diameter.
For the given equation of the circle $x\left( {x - 4} \right) + y\left( {y - 3} \right) = 0$ the points on the diameter can be found by simplifying the equation as
$\left( {x - 0} \right)\left( {x - 4} \right) + \left( {y - 0} \right)\left( {y - 3} \right) = 0$
Thus points on the diameter are $\left( {0,0} \right)$ and $\left( {4,3} \right)$.
The centre of the circle will be the mid point of the $\left( {0,0} \right)$ and $\left( {4,3} \right)$.
${C_1} = \left( {\dfrac{{0 + 4}}{2},\dfrac{{0 + 3}}{2}} \right)$
${C_1} = \left( {2,\dfrac{3}{2}} \right)$
Also, the radius of the circle is given by the distance formula
$r = \sqrt {{{\left( {2 - 0} \right)}^2} + {{\left( {\dfrac{3}{2} - 0} \right)}^2}} \\\ r = \sqrt {4 + \dfrac{9}{4}} \\\ r = \dfrac{5}{2} \\\$
The given equation for the lines of the normal is ${x^2} + 2xy + 3x + 6y = 0$.
Simplifying the equation for the lines of the normal, we get
$x\left( {x + 2y} \right) + 3\left( {x + 2y} \right) = 0 \\\ \left( {x + 3} \right)\left( {x + 2y} \right) = 0 \\\ x + 3 = 0,x + 2y = 0 \\\$
It is known that the intersection of the normal lines to the circle gives the centre of the circle.
The intersection of the normal lines $x + 3 = 0$ and $x + 2y = 0$ is
$x = - 3$
And
$2y = 3 \\\ y = \dfrac{3}{2} \\\$
Thus the centre of the required circle is ${C_2} = \left( { - 3,\dfrac{3}{2}} \right)$
Also it is known that is one circle just contains another circle then the relation ${C_1}{C_2} = R - r$ is valid, where ${C_1}$and ${C_2}$ are the centre of the two circle, and $R$is the radius of the bigger circle and $r$ is the radius of the smaller circle.
Substituting the value ${C_1} = \left( {2,\dfrac{3}{2}} \right)$, ${C_2} = \left( { - 3,\dfrac{3}{2}} \right)$ and $r = \dfrac{5}{2}$ in the equation ${C_1}{C_2} = R - r$.
$\sqrt {{{\left( {2 - \left( { - 3} \right)} \right)}^2} + {{\left( {\dfrac{3}{2} - \dfrac{3}{2}} \right)}^2}} = R - \dfrac{5}{2}$
Solving for $R$in the equation $\sqrt {{{\left( {2 - \left( { - 3} \right)} \right)}^2} + {{\left( {\dfrac{3}{2} - \dfrac{3}{2}} \right)}^2}} = R - \dfrac{5}{2}$ we get
$\sqrt {{5^2}} = R - \dfrac{5}{2} \\\ R = \dfrac{5}{2} + 5 \\\ R = \dfrac{{15}}{2} \\\$
The equation of the required circle can be found by the centre ${C_2} = \left( { - 3,\dfrac{3}{2}} \right)$ and the radius $R = \dfrac{{15}}{2}$ of the circle.
${\left( {x - \left( { - 3} \right)} \right)^2} + {\left( {y - \dfrac{3}{2}} \right)^2} = {\left( {\dfrac{{15}}{2}} \right)^2}$
Simplifying the equation, we get
${x^2} + 6x + 9 + {y^2} - 3y + \dfrac{9}{4} = \dfrac{{225}}{4} \\\ 4{x^2} + 4{y^2} + 24x - 12y - 180 = 0 \\\ {x^2} + {y^2} + 6x - 3y - 45 = 0 \\\$
Hence the equation of the required circle is ${x^2} + {y^2} + 6x - 3y - 45 = 0$.
Thus option B is the correct answer.

Note: It is known that the equation of the circle is given by the $\left( {x - {x_1}} \right)\left( {x - {x_2}} \right) + \left( {y - {y_1}} \right)\left( {y - {y_2}} \right) = 0$, where $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ are points on the extreme end of the diameter. It is known that the intersection of the normal lines to the circle gives the centre of the circle. Also it is known that is one circle just contains another circle then the relation ${C_1}{C_2} = R - r$ is valid, where ${C_1}$and ${C_2}$ are the centre of the two circle, and $R$ is the radius of the bigger circle and $r$ is the radius of the smaller circle.