Question

63. The equation of the chord of
x² + y2 - 4x +6y +3 = 0 whose mid point is 7
(1,-2) is
1) x+y+1=0
2) 2x+3y+4=0
3) x-y-3=0
4) not existing​

Answer 1

CORRECT QUESTION.

The equation of chord x² + y² - 4x + 6y + 3 = 0

whose mid point is ( 1,-2) is.

EXPLANATION.

$\sf : \implies \: equation \: of \: \: chord \\ \\ \sf : \implies \: {x}^{2} + {y}^{2} - 4x + 6y + 3 = 0 \\ \\ \sf : \implies \: centre \: of \: circle \: = ( - g , - f) = (2 , - 3) \\ \\ \sf : \implies \: radius \: = \sqrt{ {g}^{2} + {f}^{2} - c } \\ \\ \sf : \implies \: r = \sqrt{ {(2)}^{2} + ( - 3) {}^{2} - 3 } \\ \\ \sf : \implies \: r \: = \sqrt{4 + 9 - 3} = \sqrt{10}$

$\sf : \implies \: we \: can \: find \: slope \: of \: the \: equation \\ \\ \sf : \implies \: \frac{ y_{2} - y_{1} }{ x_{2} - x_{1}} = \frac{ - 3 + 2}{2 - 1} = - 1 \\ \\ \sf : \implies \: m_{1} m_{2} = - 1 \\ \\ \sf : \implies \: m _{1} ( - 1) = - 1 \\ \\ \sf : \implies \: m = 1$

$\sf : \implies \: we \: can \: find \: equation \: of \: chord \\ \\ \sf : \implies \: (y - y_{1}) = m(x - x_{1}) \\ \\ \sf : \implies \: (y + 2) = 1(x - 1) \\ \\ \sf : \implies \: y + 2 = x - 1 \\ \\ \sf : \implies \: \: x - y - 3 = 0$