A) \[{{(y-p)}^{2}}=4qx\]
B) \[{{(y-q)}^{2}}=4py\]
C) \[{{(x-p)}^{2}}=4qy\]
D) \[{{(y-q)}^{2}}=4px\]
Correct Answer: C
Solution :
Let other end of a diameter be\[(\alpha ,\beta )\]. \[\therefore \]Equation of circle is \[(x-p)(x-\alpha )+(y-q)(y-\beta )=0\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}-(p+\alpha )x-(q+\beta )y+p\alpha +q\beta =0\] This circle touch\[x-\]axis, then\[y=0\] \[\therefore \]\[{{x}^{2}}-(p+\alpha )x-(q+\beta )y+p\alpha +q\beta =0\] \[\therefore \]Discriminant, \[D=0\] \[\Rightarrow \] \[{{(p+\alpha )}^{2}}-4(p\alpha +q\beta )=0\] \[\Rightarrow \] \[{{(p-\alpha )}^{2}}=4q\beta \] Hence, locus of a point is \[{{(p-x)}^{2}}=4qy\]
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