Question

Find the equation of the circle passing through the point (-2, 4) and through the point of intersection of the circle \[{x^2} + {y^2} - 2x - 6y + 6 = 0\] and the line \[3x + 2y - 5 = 0\] .
A. \[{x^2} + {y^2} + 2x - 4y - 4 = 0\]
B. \[{x^2} + {y^2} + 4x - 2y - 4 = 0\]
C. \[{x^2} + {y^2} + 3x - 4y = 0\]
D. \[{x^2} + {y^2} - 4x - 2y - 4 = 0\]

Answer 1

Hint: Write the formula of a circle passing through the intersection of two curves. Then substitute -2 for x and 4 for y in the obtained equation to obtain the value of \[\lambda \] . Then substitute the value of \[\lambda \] in the obtained equation to obtain the required equation of the circle.

Formula used:
The equation of circle passing through the intersection of a line \[{l_1}\] and a circle \[{c_1}\] is,
\[{c_1} + \lambda {l_1} = 0\] .

Complete step by step solution:
 We know that the equation of circle passing through the intersection of a line \[{l_1}\] and a circle \[{c_1}\] is,
\[{c_1} + \lambda {l_1} = 0\] .
Now, substitute \[{x^2} + {y^2} - 2x - 6y + 6 = 0\] for \[{c_1}\] and \[3x + 2y - 5 = 0\] for \[{l_1}\] in the equation \[{c_1} + \lambda {l_1} = 0\] for further calculation.
\[\left( {{x^2} + {y^2} - 2x - 6y + 6} \right) + \lambda \left( {3x + 2y - 5} \right) = 0\]---(1)
It is given that equation (1) passes through the point \[( - 2,4)\] .
Therefore, substitute -2 for x and 4 for y in the equation \[\left( {{x^2} + {y^2} - 2x - 6y + 6} \right) + \lambda \left( {3x + 2y - 5} \right) = 0\]to obtain the value of \[\lambda \].
\[\left( {{{\left( { - 2} \right)}^2} + {{\left( 4 \right)}^2} - 2\left( { - 2} \right) - 6\left( 4 \right) + 6} \right) + \lambda \left( {3\left( { - 2} \right) + 2\left( 4 \right) - 5} \right) = 0\]
\[\left( {4 + 16 + 4 - 24 + 6} \right) + \lambda \left( { - 6 + 8 - 5} \right) = 0\]
\[6 - 3\lambda = 0\]
\[\lambda = 2\]
Substitute \[\lambda = 2\] in equation (1) to obtain the required solution.
\[\left( {{x^2} + {y^2} - 2x - 6y + 6} \right) + 2\left( {3x + 2y - 5} \right) = 0\]
\[{x^2} + {y^2} - 2x - 6y + 6 + 6x + 4y - 10 = 0\]
\[{x^2} + {y^2} + 4x - 2y - 4 = 0\]

The correct option is B.

Note: It is important to note that while substituting the given point in the equation and while finding the value of \[\lambda \] we should be avoiding any calculation mistakes and should not neglect any of the terms because it changes the value of \[\lambda \] and further the equation of the circle also.