Circles — JEE Main practice

1. Set up the tangent geometry. Let $T(h,k)$ be the point of intersection of the tangents $PQ$ and $MN$. Since $O$ is the centre and $OA\perp TA$, $OB\perp TB$, the quadrilateral $OATB$ has two right angles, so

$$\angle AOB+\angle ATB=\pi$$

Given $\angle AOB=\frac{\pi}{3}$, this gives $\angle ATB=\frac{2\pi}{3}$, so each half-angle at $T$ is

$$\angle ATO=\frac{\pi}{3}$$

2. Relate this angle to the tangent length. In right triangle $OAT$, with $OA=r$ (the radius) and $AT=L$ (the tangent length from $T$),

$$\tan(\angle ATO)=\frac{OA}{AT}=\frac{r}{L}$$

3. Find the radius of the given circle. For

$$x^2+y^2-4x-6y-3=0$$

the radius is

$$r=\sqrt{4+9+3}=4$$

4. Express the tangent length from $T(h,k)$. The length of the tangent from an external point $(h,k)$ to a circle $x^2+y^2+2gx+2fy+c=0$ is $\sqrt{h^2+k^2+2gh+2fk+c}$, so here

$$L=\sqrt{h^2+k^2-4h-6k-3}$$

5. Substitute into the relation from step 2.

$$\tan\frac{\pi}{3}=\frac{4}{\sqrt{h^2+k^2-4h-6k-3}}$$

6. Square both sides and simplify. Since $\tan\frac{\pi}{3}=\sqrt3$,

$$3\left(h^2+k^2-4h-6k-3\right)=16$$

7. Rearrange into the locus equation.

$$3h^2+3k^2-12h-18k-9-16=0 \implies 3h^2+3k^2-12h-18k-25=0$$

Replacing $(h,k)$ with $(x,y)$,

$$3(x^2+y^2)-12x-18y-25=0$$

Hence, the locus of the point of intersection of $PQ$ and $MN$ is $3(x^2+y^2)-12x-18y-25=0$, so the answer is Option C.

1. Find the centre and radius of the second circle. Given

$$x^2+y^2-4x-2y-4=0$$

completing the square,

$$(x-2)^2+(y-1)^2=9$$

so its centre is $(2,1)$ and radius $r_2=3$.

2. Identify the first circle's centre and radius. The first circle

$$(x+1)^2+(y+4)^2=r^2$$

has centre $(-1,-4)$ and radius $r_1=r$.

3. Compute the distance between the two centres.

$$c_1c_2=\sqrt{(2+1)^2+(1+4)^2}=\sqrt{9+25}=\sqrt{34}$$

4. Apply the condition for two circles to intersect at two distinct points. Two circles intersect at exactly two points when the distance between centres lies strictly between the difference and the sum of their radii:

$$|r_1-r_2|<c_1c_2<r_1+r_2$$

Substituting,

$$|r-3|<\sqrt{34}<r+3$$

5. Solve the inequality for $r$. This gives

$$r\in(\sqrt{34}-3,\ \sqrt{34}+3)$$

so $\alpha=\sqrt{34}-3$ and $\beta=\sqrt{34}+3$.

6. Compute $\alpha\beta$. Using the difference of squares,

$$\alpha\beta=(\sqrt{34})^2-3^2=34-9=25$$

Hence, the value of $\alpha\beta$ is $25$, so the answer is Option D.

1. Recognize that $AB$ is a diameter. Since $A$ lies on the $x$-axis and $B$ lies on the $y$-axis, the angle $\angle AOB=90^\circ$. An angle inscribed in a semicircle is always a right angle, so by the converse of this property, $AB$ must be the diameter of the circle.

2. Use the given radius to find $AB$. Since the radius is $4$,

$$AB=2\times4=8$$

3. Apply the distance formula to $A$ and $B$.

$$\sqrt{3a^2+2b^2}=8 \implies 3a^2+2b^2=64$$

4. Express $a$ and $b$ in terms of the centroid $(h,k)$. For triangle $OAB$ with $O=(0,0)$, $A=(-\sqrt3a,0)$, $B=(0,-\sqrt2b)$, the centroid coordinates are

$$h=\frac{-\sqrt3a}{3} \implies a=-\sqrt3\,h$$ $$k=\frac{-\sqrt2b}{3} \implies b=-\frac{3k}{\sqrt2}$$

5. Substitute these into the constraint from step 3.

$$3(3h^2)+2\left(\frac{9k^2}{2}\right)=64$$

6. Simplify.

$$9h^2+9k^2=64 \implies h^2+k^2=\frac{64}{9}$$

7. Identify the locus. Replacing $(h,k)$ with $(x,y)$,

$$x^2+y^2=\left(\frac83\right)^2$$

This is a circle of radius $\frac83$.

Hence, the radius of the locus of the centroid is $\frac{8}{3}$, so the answer is Option B.