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The equation of a circle which passes through the point (1,-2) and (4,-3) and whose centre lies on the line 3x+4y=7 is

Option 1)
$15(x^{2}+y^{2})-84x+18y-55=0$

Option 2)
$15(x^{2}+y^{2})-94x+18y+55=0$

Option 3)
$15(x^{2}+y^{2})+84x-18y+55=0$

Option 4)
None of these

Answers (1)

best_answer

Family of circle -

$S+KL= 0$

- wherein

Equation of the family of circles passing through point of intersection $S=0 \, and\, line\, L=0$ .

Equation of family of circles

$\left ( x-1 \right )\left ( x-4 \right )+\left ( y+2 \right )\left ( y+3 \right )$

$+k\begin{vmatrix} x & y & 1\\ 1 & -2& 1\\ 4& -3 & 1 \end{vmatrix} =0$

$\Rightarrow x^{2}+y^{2}-5x+5y+10+k\left ( x+3y+5 \right )=0$

center $\left ( \frac{5-k}{2}, -\frac{5+3k}{2} \right )$ his on $3x+4y =7$

$\Rightarrow \frac{3}{2}\left ( 5-k \right )-2\left ( 5+3k \right ) =7$

$\Rightarrow 15-3k -20-12k =14$

$\Rightarrow 15k = -19$

$\Rightarrow k=\frac{-19}{15}$

$x^{2}+y_{2}-5x+10 -\frac{19}{15}\left ( x+3y+5 \right )=0$

$\Rightarrow 15 \left ( x^{2}+y^{2} \right )-94x+18y+55 =0$

Option 1)

$15(x^{2}+y^{2})-84x+18y-55=0$

Incorrect

Option 2)

$15(x^{2}+y^{2})-94x+18y+55=0$

Correct

Option 3)

$15(x^{2}+y^{2})+84x-18y+55=0$

Incorrect

Option 4)

None of these

Incorrect

Posted by

divya.saini

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