JEE Main 2022 (Online) 25th July Evening ShiftQuestion 1
If the circles $x^2+y^2+6x+8y+16=0$ and $x^2+y^2+2(3-\sqrt3)x+2(4-\sqrt6)y=k+6\sqrt3+8\sqrt6$, $k>0$, touch internally at the point $P(\alpha,\beta)$, then $(\alpha+\sqrt3)^2+(\beta+\sqrt6)^2$ is equal to ________.
Circle
Answer$25$
Solution
Concept. For internal touching, $d=|r_2-r_1|$, and the point of contact is on the line of centres beyond the smaller circle's centre. Formulas used. (1) Internal touching: $C_1C_2=|r_2-r_1|$(2) $P=C_1+r_1\cdot\dfrac{C_1-C_2}{C_1C_2}$ (when $C_1$ is the centre of the smaller circle)Step 1 - First circle. Centre $C_1(-3,-4)$, $r_1^2=9+16-16=9$, $r_1=3$. Step 2 - Second circle. Centre $C_2(-3+\sqrt3,\ -4+\sqrt6)$; $$r_2^2=(3-\sqrt3)^2+(4-\sqrt6)^2+k+6\sqrt3+8\sqrt6=34+k.$$ Step 3 - Use (1). $C_1C_2=\sqrt{3+6}=3$, so $|r_2-3|=3\Rightarrow r_2=6$ ($k=2>0$ ✓). Step 4 - Use (2). $C_1-C_2=(-\sqrt3,-\sqrt6)$, length $3$: $$P=(-3,-4)+(-\sqrt3,-\sqrt6)=(-3-\sqrt3,\ -4-\sqrt6).$$ (Check: $PC_2=6$ ✓.)Step 5 - Value. $(\alpha+\sqrt3,\beta+\sqrt6)=(-3,-4)$, so the value is $9+16=25$. Answer: 25