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$x^2 + y^2 - 4x - 6y - 21 = 0$ અને $3x + 4y + 5 = 0$ ના છેદબિંદુમાંથી અને બિંદુ $(1, 2)$ માંથી પસાર થતા વર્તૂળનું સમીકરણ :
$x^2 + y^2 + 2x + 2y + 11 = 0$
$x^2 + y^2 + 2x + 2y - 11 = 0$
$x^2 + y^2 - 2x + 2y - 7 =0$
$x^2 + y^2 + 2x - 2y - 3 = 0$
Solution
The family of circles passing through intersection of a given circle $x ^2+ y ^2+$ $2 gx +2 fy + c =0$ and given line $lx + my + n =0$ is, $x ^2+ y ^2+2 gx +2 fy + c +\lambda( lx + my + n )=0$ where $\lambda \in R$
So, for given case, we can write equation as,
$x^2+y^2-4 x-6 y-21+\lambda(3 x+4 y+5)=0$
It is given that circle passes through $(1,2)$, so
$1^2+2^2-4-12-21+\lambda(3+8+5)=0$
$\rightarrow 16 \lambda=32$
$\rightarrow \lambda=2$
So, equation of cirlce,
$\rightarrow x^2+y^2-4 x-6 y-21-2(3 x+4 y+5)=0$
$\rightarrow x^2+y^2+2 x+2 y-11=0$
