Let the mirror image of a circle $c_{1}: x^{2}+y^{2}-2 x- 6 y+\alpha=0$ in line $y=x+1$ be $c_{2}: 5 x^{2}+5 y^{2}+10 g x +10 \mathrm{fy}+38=0$. If r is the radius of circle $\mathrm{c}_{2}$, then $\alpha+6 \mathrm{r}^{2}$ is equal to ______.

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Circle

Let the mirror image of a circle

c_{1} : x^{2} + y^{2} - 2 x - 6 y + α = 0

in line

y = x + 1

c_{2} : 5 x^{2} + 5 y^{2} + 10 g x + 10 fy + 38 = 0

. If r is the radius of circle

c_{2}

, then

α + 6 r^{2}

is equal to ______.

Answer: 12

Solution

Circle

c_{1}

has centre

(1, 3)

and radius

1 + 9 - α = 10 - α

.

The mirror image of

(1, 3)

in the line

x - y + 1 = 0

is found using the reflection formula:

\frac{x _{1} - 1}{1} = \frac{y _{1} - 3}{- 1} = \frac{- 2 ( 1 - 3 + 1 )}{2} = \frac{2}{2} = 1

(x_{1}, y_{1}) = (2, 2)

.

Circle

c_{2}

in standard form:

x^{2} + y^{2} - 4 x - 4 y + \frac{38}{5} = 0

, with radius

r = 4 + 4 - \frac{38}{5} = \frac{2}{5}

.

Since reflection preserves radius:

10 - α = \frac{2}{5}

, giving

α = \frac{48}{5}

.

Therefore

α + 6 r^{2} = \frac{48}{5} + 6 \cdot \frac{2}{5} = \frac{48 + 12}{5} = 12

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Integer answer

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