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Circle question

2022 · 29 Jul · Shift 1 · Q47
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  5. /2022 · 29 Jul · Shift 1 · Q47

Circle question

2022 · 29 Jul · Shift 1 · Q47

JEE MainMathematicsCircleNumerical+4 / −1
Let the mirror image of a circle c1:x2+y2−2x−6y+α=0c_{1}: x^{2}+y^{2}-2 x-6 y+\alpha=0c1​:x2+y2−2x−6y+α=0 in line y=x+1y=x+1y=x+1 be c2:5x2+5y2+10gx+10fy+38=0c_{2}: 5 x^{2}+5 y^{2}+10 g x+10 f y+38=0c2​:5x2+5y2+10gx+10fy+38=0. If r\mathrm{r}r is the radius of circle c2\mathrm{c}_{2}c2​, then α+6r2\alpha+6 \mathrm{r}^{2}α+6r2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 12

  1. Write circle c1c_1c1​ in standard form

Given x2+y2−2x−6y+α=0x^2+y^2-2x-6y+\alpha=0x2+y2−2x−6y+α=0

Compare with x2+y2−2hx−2ky+c=0x^2+y^2-2hx-2ky+c=0x2+y2−2hx−2ky+c=0 so the center is (h,k)=(1,3).(h,k)=(1,3).(h,k)=(1,3).

Also, r12=h2+k2−α=12+32−α=10−α.r_1^2=h^2+k^2-\alpha=1^2+3^2-\alpha=10-\alpha.r12​=h2+k2−α=12+32−α=10−α.

  1. Use reflection in the line y=x+1y=x+1y=x+1

The line is y=x+1  ⟺  y−x−1=0.y=x+1 \iff y-x-1=0.y=x+1⟺y−x−1=0.

Reflection preserves radius, so circle c2c_2c2​ has the same radius as c1c_1c1​. We only need the reflected center.

To reflect point (x0,y0)(x_0,y_0)(x0​,y0​) across line ax+by+c=0ax+by+c=0ax+by+c=0, image is (x0−2a(ax0+by0+c)a2+b2,  y0−2b(ax0+by0+c)a2+b2).\left(x_0-\frac{2a(ax_0+by_0+c)}{a^2+b^2},\; y_0-\frac{2b(ax_0+by_0+c)}{a^2+b^2}\right).(x0​−a2+b22a(ax0​+by0​+c)​,y0​−a2+b22b(ax0​+by0​+c)​).

Here, a=−1,b=1,c=−1a=-1,\quad b=1,\quad c=-1a=−1,b=1,c=−1 for line −x+y−1=0.-x+y-1=0.−x+y−1=0.

For center (1,3)(1,3)(1,3), ax0+by0+c=−1(1)+1(3)−1=1.ax_0+by_0+c=-1(1)+1(3)-1=1.ax0​+by0​+c=−1(1)+1(3)−1=1.

Hence reflected center is

=\left(1+1,\,3-1\right)=(2,2).$$ So $c_2$ has center $(2,2)$. 3. **Extract radius of $c_2$ from its equation** Given $$5x^2+5y^2+10gx+10fy+38=0.$$ Divide by $5$: $$x^2+y^2+2gx+2fy+\frac{38}{5}=0.$$ For a circle $$x^2+y^2+2gx+2fy+c=0,$$ center is $$(-g,-f).$$ Since center of $c_2$ is $(2,2)$, we get $$-g=2\Rightarrow g=-2, \qquad -f=2\Rightarrow f=-2.$$ Then $$r^2=g^2+f^2-\frac{38}{5}=4+4-\frac{38}{5}=\frac{40-38}{5}=\frac{2}{5}.$$ Thus, $$r^2=\frac{2}{5}.$$ 4. **Relate $\alpha$ using radius preservation** Reflection preserves radius, so $$r_1^2=r^2.$$ Thus, $$10-\alpha=\frac{2}{5}$$ which gives $$\alpha=10-\frac{2}{5}=\frac{48}{5}.$$ 5. **Compute required value** $$\alpha+6r^2=\frac{48}{5}+6\cdot\frac{2}{5} =\frac{48}{5}+\frac{12}{5}=\frac{60}{5}=12.$$ Therefore, the required integer is $$\boxed{12}.$$
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