The equation of the circle which touches the | vedclass

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Question

(d) The given normals are $x - 3y = 0,\;x - 3 = 0$ which intersect at centre whose co-ordinates are $(3, 1)$.

The given circle is ${C_1}(3,\; - 3)$

Vedclass · Accepted Answer

${x^2} + {y^2} - 6x - 2y + 1 = 0$

Vedclass · Answer

(d) The given normals are $x - 3y = 0,\;x - 3 = 0$ which intersect at centre whose co-ordinates are $(3, 1)$.

The given circle is ${C_1}(3,\; - 3)$ $\;{r_1} = 1,$

${C_2}$ is $(3, 1)$ and ${r_2} = (?)$.

If the two circles touch externally, then ${C_1}{C_2} = {r_1} + {r_2} $

$\Rightarrow 4 = 1 + {r_2}$

$\Rightarrow {r_2} = 3$

$	herefore $${(x - 3)^2} + {(y - 1)^2} = {3^2}$ or ${x^2} + {y^2} - 6x - 2y + 1 = 0$.

The equation of the circle which touches the circle ${x^2} + {y^2} - 6x + 6y + 17 = 0$ externally and to which the lines ${x^2} - 3xy - 3x + 9y = 0$ are normals, is

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