Question

Find the equation of the circle which passes through the points of intersection of circles ${x^2} + {y^2} - 2x - 6y + 6 = 0$ and ${x^2} + {y^2} + 2x - 6y + 6 = 0$ and intersects the circle ${x^2} + {y^2} + 4x + 6y + 4 = 0$ orthogonally.

Answer 1

Here, in the given question, we need to find the equation of the circle which passes through the points of intersection of circles ${x^2} + {y^2} - 2x - 6y + 6 = 0$ and ${x^2} + {y^2} + 2x - 6y + 6 = 0$ and intersects the circle ${x^2} + {y^2} + 4x + 6y + 4 = 0$ orthogonally. At first, we will find the equation of circle which passes through the point of intersection of circles ${x^2} + {y^2} - 2x - 6y + 6 = 0$ and ${x^2} + {y^2} + 2x - 6y + 6 = 0$ using ${S_1} + \lambda \left( {{S_2} - {S_1}} \right) = 0$ formula. As we know, two circles are said to be orthogonal or said to intersect orthogonally if the tangents at their point of intersection include a right angle. If two circles are given such that,
$A = {x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$
$B = {x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0$
Then they will be orthogonal only if the following condition is satisfied.
$2g{g_1} + 2f{f_1} = c + {c_1}$
As we are given that both circles are orthogonal, so using the above written formula we will find the value of $\lambda$ and substitute it in our required equation of circle.

Complete answer:
Let ${S_1} = {x^2} + {y^2} - 2x - 6y + 6 = 0$
${S_2} = {x^2} + {y^2} + 2x - 6y + 6 = 0$
Equation of circle which passes through the point of intersection of circles is given by: ${S_1} + \lambda \left( {{S_2} - {S_1}} \right) = 0$
On substituting the values, we get
$\Rightarrow \left( {{x^2} + {y^2} - 2x - 6y + 6} \right) + \lambda \left( {{x^2} + {y^2} + 2x - 6y + 6 - \left( {{x^2} + {y^2} - 2x - 6y + 6} \right)} \right) = 0$
$\Rightarrow \left( {{x^2} + {y^2} - 2x - 6y + 6} \right) + \lambda \left( {{x^2} + {y^2} + 2x - 6y + 6 - {x^2} - {y^2} + 2x + 6y - 6} \right) = 0$
On addition and subtraction of like terms, we get
$\Rightarrow {x^2} + {y^2} - 2x - 6y + 6 + \lambda \left( {4x} \right) = 0$
$\Rightarrow {x^2} + {y^2} + x\left( {4\lambda - 2} \right) - 6y + 6 = 0........\left( i \right)$
The above written equation of a circle cuts the other circle $\left( {{x^2} + {y^2} + 4x + 6y + 4 = 0} \right)$ orthogonally.
The condition of orthogonal intersection of two circles is: $2g{g_1} + 2f{f_1} = c + {c_1}$
$\Rightarrow 2\left( {2\lambda - 1} \right)2 + 2\left( { - 3} \right)3 = 6 + 4$
On multiplication of terms written outside the brackets, we get
$\Rightarrow 4\left( {2\lambda - 1} \right) + 6\left( { - 3} \right) = 10$
$\Rightarrow 8\lambda - 4 - 18 = 10$
On addition of negative terms, we get
$\Rightarrow 8\lambda - 22 = 10$
$\Rightarrow 8\lambda = 10 + 22$
On addition of terms, we get
$\Rightarrow 8\lambda = 32$
$\Rightarrow \lambda = 4$
Now, we will substitute the value of $\lambda$ in the equation $\left( i \right)$.
$\Rightarrow {x^2} + {y^2} + x\left( {4 \times 4 - 2} \right) - 6y + 6 = 0$
On multiplication of terms, we get
$\Rightarrow {x^2} + {y^2} + x\left( {16 - 2} \right) - 6y + 6 = 0$
On subtraction, we get
$\Rightarrow {x^2} + {y^2} + x\left( {14} \right) - 6y + 6 = 0$
$\Rightarrow {x^2} + {y^2} + 14x - 6y + 6 = 0$
Thus, the equation of the circle which is orthogonal to the given circle $\left( {{x^2} + {y^2} + 4x + 6y + 4 = 0} \right)$ is ${x^2} + {y^2} + 14x - 6y + 6 = 0$.

Note:
To solve this type of questions, one must remember the formulas. As we know that the general equation of family of circles passing through the intersection of ${S_1}$ and ${S_2}$ is given by ${S_1} + \lambda {S_2} = 0$ where $\lambda \ne 1$, but if we get $\lambda = 1$ we will get equation of common chord i.e. straight line instead of circle. Remember that the condition of orthogonal intersection of two circles is given by: $2g{g_1} + 2f{f_1} = c + {c_1}$. We should take care of the calculations so as to be sure of our final answer.