Let circle C be the image of x^2+y^2-2 x+4 y-4=0 in the line 2 x-3 y+5=0 and A be the point on C such that O A is parallel to x -axis and A lies on the right hand side of the centre O of C . If B(\alpha, \beta) , with \beta\lt 4 , lies on C such that the length of the arc A B is (1 / 6)^ th of the perimeter of C , then \beta-\sqrt3 \alpha is equal to

Circle PYQ — JEE Main 2025 · 24 Jan · Shift 1 · Q45 with Solution

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Correct answer: C

Find the original circle

Given circle: $x^2+y^2-2x+4y-4=0$

Complete squares: $x^2-2x+y^2+4y=4$ $(x-1)^2-1+(y+2)^2-4=4$ $(x-1)^2+(y+2)^2=9$

So the original circle has:

centre $S=(1,-2)$
radius $r=3$

Reflect the centre in the line $2x-3y+5=0$

Since reflection of a circle in a line reflects its centre and preserves radius, the image circle $C$ has radius $3$ and centre equal to the reflection of $(1,-2)$ in $2x-3y+5=0.$

For reflection of point $(x_1,y_1)$ in line $ax+by+c=0$ :

\qquad y' = y_1 - \frac{2b(ax_1+by_1+c)}{a^2+b^2}$$ Here: $$a=2,\ b=-3,\ c=5,\ (x_1,y_1)=(1,-2)$$ Compute: $$ax_1+by_1+c = 2(1)+(-3)(-2)+5 = 2+6+5=13$$ $$a^2+b^2 = 4+9=13$$ Thus $$x' = 1 - \frac{2\cdot 2\cdot 13}{13}=1-4=-3$$ $$y' = -2 - \frac{2\cdot(-3)\cdot 13}{13}=-2+6=4$$ So the centre of image circle $C$ is $$O=(-3,4)$$ with radius $3$. Hence equation of $C$ is $$(x+3)^2+(y-4)^2=9.$$ --- 3. **Find point $A$** $OA$ is parallel to the $x$-axis and $A$ lies to the right of the centre $O(-3,4)$. So $A$ is the rightmost point of the circle: $$A=(-3+3,4)=(0,4).$$ --- 4. **Use arc length condition** The perimeter (circumference) of the circle is $$2\pi r=6\pi.$$ Given arc $AB$ has length equal to $\frac16$ of the perimeter: $$\text{arc }AB = \frac16\cdot 6\pi = \pi.$$ Since arc length $s=r\theta$, $$\pi = 3\theta \implies \theta = \frac{\pi}{3}.$$ So the central angle subtended by the minor arc $AB$ is $\frac{\pi}{3}$. --- 5. **Locate point $B$** Point $A$ corresponds to angle $0$ from the positive $x$-direction at centre $O$. A point on the circle at angle $\theta$ from $OA$ has coordinates $$x=-3+3\cos\theta, \qquad y=4+3\sin\theta.$$ For $\theta=\pm \frac{\pi}{3}$, possible points are: $$B_1=\left(-3+3\cos\frac\pi3,\ 4+3\sin\frac\pi3\right) =\left(-\frac32,\ 4+\frac{3\sqrt3}{2}\right)$$ and $$B_2=\left(-3+3\cos\left(-\frac\pi3\right),\ 4+3\sin\left(-\frac\pi3\right)\right) =\left(-\frac32,\ 4-\frac{3\sqrt3}{2}\right).$$ Given $\beta<4$, we choose $$B=\left(-\frac32,\ 4-\frac{3\sqrt3}{2}\right).$$ Thus $$\alpha=-\frac32,\qquad \beta=4-\frac{3\sqrt3}{2}.$$ --- 6. **Compute $\beta-\sqrt3\alpha$** $$\beta-\sqrt3\alpha =\left(4-\frac{3\sqrt3}{2}\right)-\sqrt3\left(-\frac32\right)$$ $$=4-\frac{3\sqrt3}{2}+\frac{3\sqrt3}{2}=4.$$ --- 7. **Match with options** So, $$\beta-\sqrt3\alpha = 4.$$ Hence the correct option is: **C**.

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View written solutionFree

Correct answer: C

Find the original circle

Given circle: $x^2+y^2-2x+4y-4=0$

Complete squares: $x^2-2x+y^2+4y=4$ $(x-1)^2-1+(y+2)^2-4=4$ $(x-1)^2+(y+2)^2=9$

So the original circle has:

centre $S=(1,-2)$
radius $r=3$

Reflect the centre in the line $2x-3y+5=0$

Since reflection of a circle in a line reflects its centre and preserves radius, the image circle $C$ has radius $3$ and centre equal to the reflection of $(1,-2)$ in $2x-3y+5=0.$

For reflection of point $(x_1,y_1)$ in line $ax+by+c=0$ :

\qquad y' = y_1 - \frac{2b(ax_1+by_1+c)}{a^2+b^2}$$ Here: $$a=2,\ b=-3,\ c=5,\ (x_1,y_1)=(1,-2)$$ Compute: $$ax_1+by_1+c = 2(1)+(-3)(-2)+5 = 2+6+5=13$$ $$a^2+b^2 = 4+9=13$$ Thus $$x' = 1 - \frac{2\cdot 2\cdot 13}{13}=1-4=-3$$ $$y' = -2 - \frac{2\cdot(-3)\cdot 13}{13}=-2+6=4$$ So the centre of image circle $C$ is $$O=(-3,4)$$ with radius $3$. Hence equation of $C$ is $$(x+3)^2+(y-4)^2=9.$$ --- 3. **Find point $A$** $OA$ is parallel to the $x$-axis and $A$ lies to the right of the centre $O(-3,4)$. So $A$ is the rightmost point of the circle: $$A=(-3+3,4)=(0,4).$$ --- 4. **Use arc length condition** The perimeter (circumference) of the circle is $$2\pi r=6\pi.$$ Given arc $AB$ has length equal to $\frac16$ of the perimeter: $$\text{arc }AB = \frac16\cdot 6\pi = \pi.$$ Since arc length $s=r\theta$, $$\pi = 3\theta \implies \theta = \frac{\pi}{3}.$$ So the central angle subtended by the minor arc $AB$ is $\frac{\pi}{3}$. --- 5. **Locate point $B$** Point $A$ corresponds to angle $0$ from the positive $x$-direction at centre $O$. A point on the circle at angle $\theta$ from $OA$ has coordinates $$x=-3+3\cos\theta, \qquad y=4+3\sin\theta.$$ For $\theta=\pm \frac{\pi}{3}$, possible points are: $$B_1=\left(-3+3\cos\frac\pi3,\ 4+3\sin\frac\pi3\right) =\left(-\frac32,\ 4+\frac{3\sqrt3}{2}\right)$$ and $$B_2=\left(-3+3\cos\left(-\frac\pi3\right),\ 4+3\sin\left(-\frac\pi3\right)\right) =\left(-\frac32,\ 4-\frac{3\sqrt3}{2}\right).$$ Given $\beta<4$, we choose $$B=\left(-\frac32,\ 4-\frac{3\sqrt3}{2}\right).$$ Thus $$\alpha=-\frac32,\qquad \beta=4-\frac{3\sqrt3}{2}.$$ --- 6. **Compute $\beta-\sqrt3\alpha$** $$\beta-\sqrt3\alpha =\left(4-\frac{3\sqrt3}{2}\right)-\sqrt3\left(-\frac32\right)$$ $$=4-\frac{3\sqrt3}{2}+\frac{3\sqrt3}{2}=4.$$ --- 7. **Match with options** So, $$\beta-\sqrt3\alpha = 4.$$ Hence the correct option is: **C**.

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