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Circle question

2023 · 8 Apr · Shift 1 · Q46
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  5. /2023 · 8 Apr · Shift 1 · Q46

Circle question

2023 · 8 Apr · Shift 1 · Q46

JEE MainMathematicsCircleNumerical+4 / −1
Consider a circle C1:x2+y2−4x−2y=α−5C_{1}: x^{2}+y^{2}-4 x-2 y=\alpha-5C1​:x2+y2−4x−2y=α−5. Let its mirror image in the line y=2x+1y=2 x+1y=2x+1 be another circle C2:5x2+5y2−10fx−10gy+36=0C_{2}: 5 x^{2}+5 y^{2}-10 f x-10 g y+36=0C2​:5x2+5y2−10fx−10gy+36=0. Let rrr be the radius of C2C_{2}C2​. Then α+r\alpha+rα+r is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 2

  1. Write C1C_1C1​ in standard form

Given x2+y2−4x−2y=α−5.x^2+y^2-4x-2y=\alpha-5.x2+y2−4x−2y=α−5. Bring all terms to one side: x2+y2−4x−2y−(α−5)=0.x^2+y^2-4x-2y-(\alpha-5)=0.x2+y2−4x−2y−(α−5)=0. Compare with x2+y2−2hx−2ky+(h2+k2−r2)=0.x^2+y^2-2hx-2ky+(h^2+k^2-r^2)=0.x2+y2−2hx−2ky+(h2+k2−r2)=0. So the center of C1C_1C1​ is (2,1).(2,1).(2,1).

Now complete squares: x2−4x+y2−2y=α−5x^2-4x+y^2-2y=\alpha-5x2−4x+y2−2y=α−5 (x−2)2−4+(y−1)2−1=α−5 (x-2)^2-4+(y-1)^2-1=\alpha-5(x−2)2−4+(y−1)2−1=α−5 (x−2)2+(y−1)2=α. (x-2)^2+(y-1)^2=\alpha.(x−2)2+(y−1)2=α. Hence radius of C1C_1C1​ is α.\sqrt{\alpha}.α​.

  1. Use the fact that mirror image of a circle has same radius

Reflection preserves distances, so the mirror image circle C2C_2C2​ has the same radius as C1C_1C1​. Thus radius of C2C_2C2​ is also r=α.r=\sqrt{\alpha}.r=α​.

  1. Find center of C2C_2C2​ from its equation

Given 5x2+5y2−10fx−10gy+36=0.5x^2+5y^2-10fx-10gy+36=0.5x2+5y2−10fx−10gy+36=0. Divide by 555: x2+y2−2fx−2gy+365=0.x^2+y^2-2fx-2gy+\frac{36}{5}=0.x2+y2−2fx−2gy+536​=0. So the center of C2C_2C2​ is (f,g),(f,g),(f,g), and its radius satisfies r2=f2+g2−365.r^2=f^2+g^2-\frac{36}{5}. r2=f2+g2−536​.

  1. Reflect the center (2,1)(2,1)(2,1) in the line y=2x+1y=2x+1y=2x+1

Write the line as 2x−y+1=0.2x-y+1=0.2x−y+1=0. For reflection of point (x0,y0)(x_0,y_0)(x0​,y0​) in 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=2,b=−1,c=1,(x0,y0)=(2,1).a=2,\quad b=-1,\quad c=1,\quad (x_0,y_0)=(2,1).a=2,b=−1,c=1,(x0​,y0​)=(2,1). Compute ax0+by0+c=2⋅2+(−1)⋅1+1=4.ax_0+by_0+c=2\cdot2+(-1)\cdot1+1=4.ax0​+by0​+c=2⋅2+(−1)⋅1+1=4. Also, a2+b2=4+1=5.a^2+b^2=4+1=5.a2+b2=4+1=5. Therefore reflected point is

=\left(2-\frac{16}{5},\;1+\frac{8}{5}\right) =\left(-\frac{6}{5},\;\frac{13}{5}\right).$$ So $$f=-\frac65,\quad g=\frac{13}{5}.$$ --- 5. **Find the radius of $C_2$** Using $$r^2=f^2+g^2-\frac{36}{5},$$ we get $$r^2=\left(-\frac65\right)^2+\left(\frac{13}{5}\right)^2-\frac{36}{5} =\frac{36}{25}+\frac{169}{25}-\frac{180}{25} =\frac{25}{25}=1.$$ So $$r=1.$$ --- 6. **Find $\alpha$** Since reflection preserves radius, $$\sqrt{\alpha}=r=1.$$ Hence $$\alpha=1.$$ --- 7. **Compute $\alpha+r$** $$\alpha+r=1+1=2.$$ ## Final Answer $$\boxed{2}$$ The derived answer matches the stored correct answer.
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