A) Intersect
B) Are concentric
C) Touch internally
D) Touch externally
Correct Answer: C
Solution :
Given, equations of the circles \[{{x}^{2}}+{{y}^{2}}-2x+6y+6\]=0 .....(i) and\[{{x}^{2}}+{{y}^{2}}-5x+6y+15=0\] .....(ii) We know that the standard equation of a circle is \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0.\] Therefore for circle (i), \[g=-1;\,f=3;\,\,\,c=6;\] centre \[A=(1,\,-3)\] and radius\[({{r}_{1}})=\sqrt{{{g}^{2}}+{{f}^{2}}-c}=\sqrt{1+9-6}=2\]. Similarly, for circle (ii), \[g=\frac{-5}{2};\,\,f=3;\,c=15;\] \ Centre \[B\equiv \,\left( +\frac{5}{2},-3 \right)\] and radius \[({{r}_{2}})=\sqrt{\frac{25}{4}+9-15}=\frac{1}{2}\] Therefore distance between A and B \[=\sqrt{{{\left( \frac{5}{2}-1 \right)}^{2}}+{{(-3+3)}^{2}}}=\frac{3}{2}\] and difference of radii \[({{r}_{1}}-{{r}_{2}})=2-\frac{1}{2}=\frac{3}{2}.\] Since distance between A and B is equal to \[{{r}_{1}}-{{r}_{2}},\] therefore the circles touch each other internally.
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