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 \[(7,\,\,-5)\].You need to login to perform this action.
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