The equation of the image of the circle {x^2} | vedclass

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Question

(d) The centre of the required circle is the image of the centre $( - 8,\;12)$ with respect to the line mirror $4x + 7y + 13 = 0$ and radius is equal

Vedclass · Accepted Answer

${x^2} + {y^2} + 32x + 4y + 235 = 0$

Vedclass · Answer

(d) The centre of the required circle is the image of the centre $( - 8,\;12)$ with respect to the line mirror $4x + 7y + 13 = 0$ and radius is equal to the radius of the given circle.
Let $(h, k)$ be the image of the point $( - 8,\;12)$ with respect to the line mirror. Then the mid-point of the line joining $C( - 8,\;12)$ and $P(h,\;k)$ lies on the line mirror.
$	herefore \;4\left( {\frac{{h - 8}}{2}} ight) + 7\left( {\frac{{k + 12}}{2}} ight) + 13 = 0$
or $4h + 7k + 78 = 0$&hellip;.(i)
Also $CP$ is perpendicular to $4x + 7y + 13 = 0$
$	herefore $ $\frac{{k - 12}}{{h + 8}} 	imes - \frac{4}{7} = - 1$ or $7h - 4k + 104 = 0$&hellip;.(ii)
Solving (i) and (ii), $h = - 16,\;k = - 2$.
Thus the centre of the image circle is $( - 16,\; - 2)$. The radius of the image circle is same as the radius of ${x^2} + {y^2} + 16x - 24y + 183 = 0$ i.e., $5$.
Hence the equation of the required circle is
${(x + 16)^2} + {(y + 2)^2} = {5^2}$ i.e. ${x^2} + {y^2} + 32x + 4y + 235 = 0$.

The equation of the image of the circle ${x^2} + {y^2} + 16x - 24y + 183 = 0$ by the line mirror $4x + 7y + 13 = 0$ is

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