question_answer

Two circles with equations \[{{x}^{2}}+{{y}^{2}}+4x-10y+13=0\] and \[{{x}^{2}}+{{y}^{2}}-12x+2y+1=0\] share one common internal tangent. The y-intercept of the common tangent is

A) 1 

B) \[\frac{4}{3}\]

C) \[\frac{2}{3}\]                                   

D) \[-2\]

Correct Answer: A

Solution :

\[{{C}_{1}}:{{r}_{1}}=\sqrt{4+25-13}=4\]                \[{{C}_{2}}:{{r}_{2}}=\sqrt{36+1-1}=6\]                 Centre (-2, 5) and (6, -1) \[d=\sqrt{64+36}=10\]                   Hence circles touch externally.                 Equation of common internal tangent is \[{{S}_{1}}-{{S}_{2}}=0\] \[16x-12y+12=0\] \[4x+3y+3=0\] y-intercept \[x=0\] \[\Rightarrow y=1\]