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Circle question

2024 · 30 Jan · Shift 2 · Q54
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  5. /2024 · 30 Jan · Shift 2 · Q54

Circle question

2024 · 30 Jan · Shift 2 · Q54

JEE MainMathematicsCircleNumerical+4 / −1
Consider two circles C1:x2+y2=25C_1: x^2+y^2=25C1​:x2+y2=25 and C2:(x−α)2+y2=16C_2:(x-\alpha)^2+y^2=16C2​:(x−α)2+y2=16, where α∈(5,9)\alpha \in(5,9)α∈(5,9). Let the angle between the two radii (one to each circle) drawn from one of the intersection points of C1C_1C1​ and C2C_2C2​ be sin⁡−1(638)\sin ^{-1}\left(\frac{\sqrt{63}}{8}\right)sin−1(863​​). If the length of common chord of C1C_1C1​ and C2C_2C2​ is β\betaβ, then the value of (αβ)2(\alpha \beta)^2(αβ)2 equals ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 1575

  1. Identify the geometry of the two circles

The circles are C1:x2+y2=25C_1:x^2+y^2=25C1​:x2+y2=25 with center O1=(0,0)O_1=(0,0)O1​=(0,0) and radius r1=5r_1=5r1​=5, and C2:(x−α)2+y2=16C_2:(x-\alpha)^2+y^2=16C2​:(x−α)2+y2=16 with center O2=(α,0)O_2=(\alpha,0)O2​=(α,0) and radius r2=4r_2=4r2​=4.

Since α∈(5,9)\alpha\in(5,9)α∈(5,9), the distance between centers is O1O2=α.O_1O_2=\alpha.O1​O2​=α.

If the circles intersect at a point PPP, then in triangle O1PO2O_1PO_2O1​PO2​:

  • O1P=5O_1P=5O1​P=5
  • O2P=4O_2P=4O2​P=4
  • O1O2=αO_1O_2=\alphaO1​O2​=α

The angle between the radii drawn to the intersection point is ∠O1PO2=sin⁡−1(638).\angle O_1PO_2=\sin^{-1}\left(\frac{\sqrt{63}}{8}\right).∠O1​PO2​=sin−1(863​​).

Let this angle be θ\thetaθ. Then sin⁡θ=638.\sin\theta=\frac{\sqrt{63}}{8}.sinθ=863​​.

  1. Find cos⁡θ\cos\thetacosθ

Since θ\thetaθ is an angle of a triangle, 0<θ<π0<\theta<\pi0<θ<π, and here the circles intersect with α∈(5,9)\alpha\in(5,9)α∈(5,9), so the angle is acute. Thus,

=\sqrt{1-\frac{63}{64}} =\frac{1}{8}.$$ 3. **Use the cosine law in triangle $O_1PO_2$ to find $\alpha$** By the cosine law, $$\alpha^2=5^2+4^2-2\cdot 5\cdot 4\cos\theta.$$ So, $$\alpha^2=25+16-40\cdot \frac18=41-5=36.$$ Hence, $$\alpha=6$$ (because $\alpha\in(5,9)$). 4. **Find the common chord length $\beta$** Let the common chord be perpendicular to the line joining centers. Since the centers lie on the $x$-axis, subtract the equations of the circles: $$x^2+y^2-\big((x-\alpha)^2+y^2\big)=25-16$$ $$x^2-(x^2-2\alpha x+\alpha^2)=9$$ $$2\alpha x-\alpha^2=9$$ $$x=\frac{\alpha^2+9}{2\alpha}.$$ For $\alpha=6$, $$x=\frac{36+9}{12}=\frac{45}{12}=\frac{15}{4}.$$ Now on circle $C_1$, the chord endpoints satisfy $$x^2+y^2=25.$$ Thus, $$y^2=25-\left(\frac{15}{4}\right)^2 =25-\frac{225}{16} =\frac{400-225}{16} =\frac{175}{16}.$$ So, $$|y|=\frac{5\sqrt7}{4}.$$ Therefore the common chord length is $$\beta=2|y|=\frac{5\sqrt7}{2}.$$ 5. **Compute $(\alpha\beta)^2$** $$\alpha\beta=6\cdot \frac{5\sqrt7}{2}=15\sqrt7.$$ Hence, $$(\alpha\beta)^2=(15\sqrt7)^2=225\cdot 7=1575.$$ 6. **Compare with stored answer** Our derived answer is $$\boxed{1575}$$ which matches the stored correct answer.
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