IIT-JEE 1993Question 1
Find the coordinates of the point at which the circles $x^{2}+y^{2}-4x-2y=-4$ and $x^{2}+y^{2}-12x-8y=-36$ touch each other. Also find equations of common tangents touching the circles in the distinct points.
Circle
AnswerPoint of contact $\left(\dfrac{14}{5},\dfrac{8}{5}\right)$; tangents $y=0$ and $7y-24x+16=0$
Solution
Concept. Check the type of contact, find the contact point on the line of centres, and get the direct tangents through the external centre of similitude. Formulas used. (1) External contact point divides $C_1C_2$ internally in $r_1:r_2$.(2) External centre of similitude divides $C_1C_2$ externally in $r_1:r_2$.(3) Distance from centre $=$ radius for a tangent.Step 1 - Circles. $C_1=(2,1)$, $r_1=\sqrt{4+1-4}=1$. $C_2=(6,4)$, $r_2=\sqrt{36+16-36}=4$. $d=5=r_1+r_2$: external contact. Step 2 - Use (1). $\frac{4(2,1)+1(6,4)}{5}=\left(\frac{14}5,\frac85\right)$. Step 3 - Use (2). $\frac{1\cdot(6,4)-4\cdot(2,1)}{1-4}=\left(\frac23,0\right)$.Step 4 - Tangents, use (3). $y=m\left(x-\frac23\right)$ at distance $1$ from $(2,1)$: $$\left(\tfrac{4m}3-1\right)^2=1+m^2\Rightarrow\tfrac{7m^2}9=\tfrac{8m}3\Rightarrow m=0\text{ or }\tfrac{24}7.$$ Step 5 - Lines. $m=0$: $y=0$. $m=\frac{24}7$: $7y=24x-16$, i.e. $7y-24x+16=0$. (Check with $C_2$: $\frac{|28-144+16|}{25}=4$.) Answer: Point of contact $\left(\frac{14}{5},\frac{8}{5}\right)$; tangents $y=0$ and $7y-24x+16=0$.