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  • A circle touches the parabola at <img alt="\mathrm{P}" src="https://entrancec

A circle touches the parabola \mathrm{y^2=4 a x} at \mathrm{P}. It also passes through the focus \mathrm{ S} of the parabola and intersects its axis at \mathrm{ \mathrm{Q}}. If angle \mathrm{ \mathrm{SPQ}} is \mathrm{ 90^{\circ}}, find the equation of the circle.

Option: 1

\mathrm{x^2+y^2-10 a x+5 a y+9 a^2=0}


Option: 2

\mathrm{x^2+y^2-10 a x+9 a^2=0}


Option: 3

\mathrm{x^2+y^2+10 a x-9 a^2=0}


Option: 4

\mathrm{2 x^2+2 y^2+10 a x+5 a y+9 a^2=0}


Answers (1)

best_answer

Let the point Q be \mathrm{(\alpha, 0).} Also S is \mathrm{(a, 0).} Since \mathrm{\angle S P Q=90^{\circ}, S Q} is a diameter of the circle.

Hence its equation is\mathrm{ (x-a)(x-a)+y^2=0}

Since it touches the \mathrm{ y^2=4 a x,}

\mathrm{(x-a)(x-a)+4 a x=0} has equal roots

i.e. \mathrm{x^2-x(a-3 a)+a \alpha=0} has equal roots

\mathrm{\Rightarrow(\alpha-3 a)^2-4 a \alpha=0 \, \, or \, \, \alpha^2-10 a \alpha+9 a^2=0}

or \mathrm{ (\alpha-a)(\alpha-9 a)=0 \Rightarrow \alpha=9 a (since \alpha \neq a).}

Hence the equation of the circle is

\mathrm{ \begin{aligned} & (x-9 a)(x-a)+y^2=0 \\\\ & \text { i.e. } x^2+y^2-10 a x+9 a^2=0 \end{aligned} }

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