question_answer

The lines joining the points of intersection of the curve \[{{(x-h)}^{2}}+{{(y-k)}^{2}}-{{c}^{2}}=0\] and the line \[kx+hy=2hk\]  to the origin are perpendicular, then

A)            \[c=h\pm k\]                       

B)            \[{{c}^{2}}={{h}^{2}}+{{k}^{2}}\]

C)            \[{{c}^{2}}={{(h+k)}^{2}}\]     

D)            \[4{{c}^{2}}={{h}^{2}}+{{k}^{2}}\]

Correct Answer: B

Solution :

           The line is \[\frac{x}{2h}+\frac{y}{2k}=1\]and circle is,            \[{{x}^{2}}+{{y}^{2}}-2(hx+ky)+({{h}^{2}}+{{k}^{2}}-{{c}^{2}})=0\]            Making it homogeneous, we get            \[\Rightarrow ({{x}^{2}}+{{y}^{2}})-2(hx+ky)\left( \frac{x}{2h}+\frac{y}{2k} \right)+\]                          If these lines be perpendicular, then \[A+B=0\]                    \[\left[ 1-1+\frac{({{h}^{2}}+{{k}^{2}}-{{c}^{2}})}{4{{h}^{2}}} \right]+\left[ 1-1+\frac{({{h}^{2}}+{{k}^{2}}-{{c}^{2}})}{4{{k}^{2}}} \right]=0\]                    or \[({{h}^{2}}+{{k}^{2}}-{{c}^{2}})\text{ }\left( \frac{{{h}^{2}}+{{k}^{2}}}{4{{h}^{2}}{{k}^{2}}} \right)=0\]                    \[\therefore {{h}^{2}}+{{k}^{2}}={{c}^{2}}\].