JEE Main 2023 — Circle Question with Solution

We are given two circles:

(1) $x^2+y^2-18x-15y+131=0$

(2) $x^2+y^2-6x-6y-7=0$

First, let's find the centers and radii of the circles.

For circle (1):

Completing the square for the equation:

$(x^2-18x+81)+(y^2-15y+\frac{225}{4})=-131+81+\frac{225}{4}$

$(x-9{)}^2+(y-\frac{15}{2}{)}^2=\frac{25}{4}$

Center 1: $C_1(9,\frac{15}{2})$

Radius 1: $r_1=\sqrt{\frac{25}{4}}=\frac{5}{2}$

For circle (2):

Completing the square for the equation:

$(x^2-6x+9)+(y^2-6y+9)=7+9+9$

$(x-3{)}^2+(y-3{)}^2=25$

Center 2: $C_2(3,3)$

Radius 2: $r_2=5$

Now, let's find the distance between the centers :

$d=\sqrt{(9-3{)}^2+(\frac{15}{2}-3{)}^2}=\sqrt{6^2+{\frac{9}{2}}^2}=\sqrt{36+\frac{81}{4}}=\frac{15}{2}$

Next, let's analyze the relative positions of the circles using the distance between centers and the sum and difference of the radii :

1. If $d>r_1+r_2$, the circles are separate, and there are 4 common tangents.
2. If $d=r_1+r_2$, the circles are externally tangent, and there are 3 common tangents.
3. If $0<d<|r_1-r_2|$, one circle is inside the other, and there are no common tangents.
4. If $d=|r_1-r_2|$, the circles are internally tangent, and there is 1 common tangent.
5. If $d<|r_1-r_2|$, one circle is completely inside the other, and there are no common tangents.

In this case :

$d=\frac{15}{2}$

$r_1=\frac{5}{2}$

$r_2=5$

Now, let's check the conditions :

$r_1+r_2=\frac{5}{2}+5$ = $\frac{15}{2}$

Since $d=\frac{15}{2}=r_1+r_2$, the circles touch each other externally, and there are 3 common tangents.