question_answer

The equation of the circle which cuts each of the three circles x2 + y2 = 4, (x-1)2 + y2 = 4 and x2 + (y - 2)2 = 4 orthogonally is

A)  \[{{x}^{2}}+{{y}^{2}}+x+2y+4=0\]                       

B)  \[{{x}^{2}}+{{y}^{2}}+x-2y+4=0\]

C)  \[{{x}^{2}}+{{y}^{2}}-x-2y+4=0~\]

D)  \[{{x}^{2}}+{{y}^{2}}-x+2y+4=0\]

Correct Answer: C

Solution :

 Idea If two circles are intersect orthogonally as figure. Then \[{{C}_{1}}C_{2}^{2}={{C}_{1}}{{P}^{2}}+{{C}_{2}}{{P}^{2}}\]Apply this condition to solve the question Given, x2 + y2 = 4 (x -1)2 + y2 = 4 and x2 + (y - 2)2 = 4 Let the required circle be \[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\] Now, by condition of orthogonality,                 \[{{h}^{2}}+{{k}^{2}}={{r}^{2}}+4\]                                           ?(i) \[\Rightarrow \]               \[{{(h-1)}^{2}}+{{k}^{2}}={{r}^{2}}+4\]                   ?(ii) and        \[{{h}^{2}}+{{(k-2)}^{2}}={{r}^{2}}+4\]                   ?(iii) \[\Rightarrow \]               \[h=\frac{1}{2}\]and\[k=1\] \[\therefore \]  \[{{r}^{2}}+4=\frac{1}{4}+1\]                      [From Eq. (i)] \[\Rightarrow \]               \[{{r}^{2}}=-\frac{11}{4}\] \[\therefore \]Required equation of circle \[{{\left( x-\frac{1}{2} \right)}^{2}}+{{(y-1)}^{2}}=-\frac{11}{4}\] \[\Rightarrow \]               \[{{x}^{2}}+{{y}^{2}}-x-2y+4=0\] So, correct option is [c]. TEST Edge Equation of circles in different cases and relation between the circle and conic section related questions are asked. To solve such types of questions students are advised to understand the concept of circle and conic section and how to circle and conic section are related with each other.