A) \[{{x}^{2}}+{{y}^{2}}+2x=2y+11=0\]
B) \[{{x}^{2}}+{{y}^{2}}-2x+2y-7=0\]
C) \[{{x}^{2}}+{{y}^{2}}-2x-2y-3=0\]
D) \[{{x}^{2}}+{{y}^{2}}+2x+2y-11=0\]
Correct Answer: D
Solution :
Point (1,2) lies on the circle \[{{x}^{2}}+{{y}^{2}}+2x+2y\]\[-11=0,\] because coordinates of point (1,2) satisfy the equation\[{{x}^{2}}+{{y}^{2}}+2x+2y-11=0\] Now,\[{{x}^{2}}+{{y}^{2}}-4x-6y-21=0\] ?(i) \[{{x}^{2}}+{{y}^{2}}+2x+2y-11=0\] ?(ii) \[3x+4y+5=0\] ?(iii) From (i) and (iii), \[{{x}^{2}}+{{\left( -\frac{3x+5}{4} \right)}^{2}}-4x-6\left( \frac{3x+5}{4} \right)-21=0\] \[\Rightarrow \]
\[+72x+120-336=0\] \[\Rightarrow \]\[25{{x}^{2}}+38x-191=0\] ?.(iv) From (ii) and (iii), \[{{x}^{2}}+{{\left( \frac{3x+5}{4} \right)}^{2}}+2x+2\left( -\frac{3x+5}{4} \right)-11=0\] \[\Rightarrow \]
\[-24x-40-176=0\] \[\Rightarrow \]\[25{{x}^{2}}+38x-191=0\] ?(v) Thus we get the same equation from (ii) and (iii) as we get from equation (i) and (iii). Hence the point of intersections of (ii) and (iii) will be same as the point of intersections of (i) and (iii). Therefore the circle (ii) passing through the point of intersection of circlet) and point (1, 2) also as shown in the figure 
Hence equation(ii) i.e. 
is the equation of required circle.
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