question_answer

The other end of the diameter through the point \[(-1,1)\]     on     the     circle \[{{x}^{2}}+{{y}^{2}}-6x+4y-12=0\]'              

A)  \[(-7,5)\]       

B)  \[(-7,-5)\]

C)  \[(7,-5)\]

D)  \[(7,5)\]

Correct Answer: C

Solution :

Given circle is \[{{x}^{2}}+{{y}^{2}}-6x+4y-12=0\] Centre of this circle is \[(3,-2).\] Let other end of the diameter is \[(\alpha ,\beta )\]. \[\therefore \] \[\frac{\alpha -1}{2}=3,\,\,\frac{\beta +1}{2}=-2\] \[\Rightarrow \] \[\alpha =7,\,\,\,\beta =-5\] \[\therefore \] Other end of the diameter is \[(7,\,\,-5)\].