The equation of a circle that intersects the | vedclass

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Question

(d) In circle, ${x^2} + {y^2} + 14x + 6y + 2 = 0$

$g = 7,\;f = 3,\;c = 2$

Centre of circle $( - g,\; - f) = (0,\;2)$, (Given)

For orthogonall

Vedclass · Accepted Answer

${x^2} + {y^2} - 4y - 14 = 0$

Vedclass · Answer

(d) In circle, ${x^2} + {y^2} + 14x + 6y + 2 = 0$

$g = 7,\;f = 3,\;c = 2$

Centre of circle $( - g,\; - f) = (0,\;2)$, (Given)

For orthogonally intersection, $2gg' + 2ff' = c + c'$

$0 - 12 = 2 + c' \Rightarrow c' = - 14$

Put the values, in equation ${x^2} + {y^2} + 2g'x + 2f'x + c' = 0$.

$ \Rightarrow {x^2} + {y^2} + 0 - 4y - 14 = 0 $

$\Rightarrow {x^2} + {y^2} - 4y - 14 = 0$.

The equation of a circle that intersects the circle ${x^2} + {y^2} + 14x + 6y + 2 = 0$orthogonally and whose centre is $(0, 2)$ is

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