The straight line {x} \cos \alpha+ {y} \sin \alpha= {p} cuts the circle {x}^{2}+ {y}^{2}- {a}^{2}=0 at {A} and {B}. Then the equation of circle having {AB} as diameter is

Correct Answer is : x2+y2−a2−2p(xcosα+ysinα−p)=0

x^{2}+y^{2}-a^{2}+p(x \cos \alpha+y \sin \alpha-p)=0

x^{2}+y^{2}-a^{2}-p(x \cos \alpha+y \sin \alpha+p)=0

x^{2}+y^{2}-a^{2}+2 p(x \cos \alpha+y \sin \alpha-p)=0

x^{2}+y^{2}-a^{2}-2 p(x \cos \alpha+y \sin \alpha-p)=0

AP EAPCET 08-Jul-2022 Shift 1 Paper

Section: Mathematics

Question : 51 of 160

Marks: +1, -0

The straight line x‌cos‌α+y‌sin‌α=p cuts the circle x2+y2−a2=0 at A and B. Then the equation of circle having AB as diameter is

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