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10-1.Circle and System of Circles
medium
The equation of the circle passing through the point $(-2, 4)$ and through the points of intersection of the circle ${x^2} + {y^2} - 2x - 6y + 6 = 0$ and the line $3x + 2y - 5 = 0$, is
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
${x^2} + {y^2} - 4x - 2y = 0$
Solution
(b) Required equation of the circle,
$({x^2} + {y^2} – 2x – 6y + 6) + \lambda (3x + 2y – 5) = 0$
This circle passing through points $( – 2,\;4)$, therefore
$(4 + 16 + 4 – 24 + 6) + \lambda ( – 6 + 8 – 5) = 0$,
$\therefore \;\lambda = 2$
$\therefore \;({x^2} + {y^2} – 2x – 6y + 6) + 2(3x + 2y – 5) = 0$
$ \Rightarrow {x^2} + {y^2} + 4x – 2y – 4 = 0$.
Standard 11
Mathematics