If the equation of the tangent to the circle | vedclass

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Question

(a) Equation of circle is,

${x^2} + {y^2} - 2x + 6y - 6 = 0$

${(x - 1)^2} + {(y + 3)^2} = {(4)^2}$

Radius of circle = $4$

And centre of ci

Vedclass · Accepted Answer

$5, -35$

Vedclass · Answer

(a) Equation of circle is,

${x^2} + {y^2} - 2x + 6y - 6 = 0$

${(x - 1)^2} + {(y + 3)^2} = {(4)^2}$

Radius of circle = $4$

And centre of circle $ = (1, - 3)$

Equation of tangent $3x - 4y + k = 0$

 $\frac{{3 	imes 1 - 4 	imes ( - 3) + k = 0}}{{\sqrt {{{(3)}^2} + {{( - 4)}^2}} }}= \pm 4$.

Hence, $k = 5, - 35$.

If the equation of the tangent to the circle ${x^2} + {y^2} - 2x + 6y - 6 = 0$ parallel to $3x - 4y + 7 = 0$ is $3x - 4y + k = 0$, then the values of $k$ are

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