Find the equation of the circle whose centre is the point of intersection of the lines $2 x - 3 y + 4 = 0$ and $3 x + 4 y - 5 = 0$ and passes through the origin.

17 (x^{2} + y^{2}) + 2 x - 44 y = 0

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17 (x^{2} - y^{2}) + 2 x + 44 y = 0

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17 (x^{2} + y^{2}) - 2 x + 44 y = 0

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Solution

The correct option is A $17 (x^{2} + y^{2}) + 2 x - 44 y = 0$
The point of intersection of the lines $2 x - 3 y + 4 = 0$ and $3 x + 4 y - 5 = 0$ is $(- \frac{1}{17}, \frac{22}{17})$
Equation of circle whose centre is $(- \frac{1}{17}, \frac{22}{17})$ and passing through origin is
$x^{2} + y^{2} + \frac{2 x}{17} - \frac{44 y}{17} = 0$
$\Rightarrow 17 (x^{2} + y^{2}) + 2 x - 44 y = 0$
Hence, option A.

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Find the equation of the circle whose centre is the point of intersection of the lines 2x−3y+4=0 and 3x+4y−5=0 and passes through the origin.

The correct option is A 17(x2+y2)+2x−44y=0The point of intersection of the lines 2x−3y+4=0 and 3x+4y−5=0 is (−117,2217)Equation of circle whose centre is (−117,2217) and passing through origin isx2+y2+2x17−44y17=0⇒17(x2+y2)+2x−44y=0Hence, option A.

Find the equation of the circle whose centre is the point of intersection of the lines $2 x - 3 y + 4 = 0$ and $3 x + 4 y - 5 = 0$ and passes through the origin.