• # question_answer A variable circle passes through the fixed point $A(p,\text{ }q)$and touches x-axis. The locus of the other end of the diameter through A is A) ${{(x-p)}^{2}}=4qy$      B) ${{(x-q)}^{2}}=4qy$ C)                        ${{(y-p)}^{2}}=4qx$      D) ${{(y-q)}^{2}}=4px$

If$A({{x}_{1}},{{y}_{1}})$and$B({{x}_{2}},{{y}_{2}})$are the coordinates of end points diameter of a circle, then the equation of circle is $({{x}_{1}},{{y}_{1}})(x-{{x}_{2}})+(y-{{y}_{1}})(y-{{y}_{2}})=0$ In a circle AB is as a diameter where the coordinates of A are$(p,\text{ }q)$and let the coordinates of $({{x}_{1}},{{y}_{1}})$. Equation of circle in diameter form is $(x-p)(x-{{x}_{1}})+(y-p)(y-{{y}_{1}})=0$ $\Rightarrow$${{x}^{2}}-(p+{{x}_{1}})x+p{{x}_{1}}+{{y}^{2}}$                                 $-({{y}_{1}}+q)y+q{{y}_{1}}=0$ $\Rightarrow$${{x}^{2}}-(p+{{x}_{1}})x+{{y}^{2}}$ $-({{y}_{1}}+q)y+p{{x}_{1}}+q{{y}_{1}}=0$ Since, this circle touches X-axis. $\therefore$  $y=0$ $\Rightarrow$ ${{x}^{2}}-(p+{{x}_{1}})x+p{{x}_{1}}+q{{y}_{1}}=0$ Also, the discriminant of above equation will be equal to zero because circle touches X-axis. $\therefore$  ${{(p+{{x}_{1}})}^{2}}=4(p{{x}_{1}}+q{{y}_{1}})$ $\Rightarrow$               ${{p}^{2}}+x_{1}^{2}+2p{{x}_{1}}=4p{{x}_{1}}+4q{{y}_{1}}$ $\Rightarrow$               $x_{1}^{2}-2p{{x}_{1}}+{{p}^{2}}=4q{{y}_{1}}$ Hence, the locus of point S is ${{(x-p)}^{2}}=4qy$