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Please help! 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

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)
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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

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