JEE Main 2019 (Online) 9th April Evening ShiftQuestion 1
The common tangent to the circles $x^{2}+y^{2}=4$ and $x^{2}+y^{2}+6x+8y-24=0$ also passes through the point :
Circle
Solution
Formulas used. (1) Circles touch internally $\iff d=|r_1-r_2|$.(2) For two touching circles, the common tangent at the point of contact is $S_1-S_2=0$.Step 1 - Centres and radii. $x^2+y^2=4$: centre $O(0,0)$, $r_1=2$. $x^2+y^2+6x+8y-24=0$: centre $(-3,-4)$, $r_2=\sqrt{9+16+24}=7$. Step 2 - Position. $d=\sqrt{9+16}=5=7-2=r_2-r_1$, so by (1) the circles touch internally. Hence they have exactly one common tangent, at the point of contact.Step 3 - The common tangent. By (2): $$(x^2+y^2-4)-(x^2+y^2+6x+8y-24)=0\Rightarrow -6x-8y+20=0$$ $$\Rightarrow 3x+4y=10$$ Step 4 - Check the options. $(4,-2)$: $12-8=4$; $(-4,6)$: $-12+24=12$; $(6,-2)$: $18-8=10$ ✓; $(-6,4)$: $-18+16=-2$. Answer: C