A line is drawn through the point P(3, 11) to cut the circle x2 + y2 = 9 at A and B. Then PA ⋅ PB is equal to

Question

A line is drawn through the point P(3, 11) to cut the circle x2 + y2 = 9 at A and B. Then PA ⋅ PB is equal to (A) 9 (B) 121 (C) 205 (D) 139

Tardigrade Admin · Accepted Answer

Given circle, x2 + y2 = 9 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/8ac56ab0be4313aaff34c6876860d26c-.png" /> PA× PB = PT 2 [ By Geometry] PA × PB = (3)2 + (11)2 - 9 = 121

Q. A line is drawn through the point P(3, 11) to cut the circle $x^{2} + y^{2} = 9$ at A and B. Then $P A \cdot PB$ is equal to

9

121

205

139

Given circle, $x^{2} + y^{2} = 9$

$P A \times PB = P T^{2}$ [ By Geometry]
$P A \times PB = (3)^{2} + (11)^{2} - 9 = 121$

Q. A line is drawn through the point P(3, 11) to cut the circle x2+y2=9 at A and B. Then PA⋅PB is equal to

9

121

205

139

Given circle, x2+y2=9 PA×PB=PT2 [ By Geometry] PA×PB=(3)2+(11)2−9=121

Q. A line is drawn through the point P(3, 11) to cut the circle $x^{2} + y^{2} = 9$ at A and B. Then $P A \cdot PB$ is equal to

Given circle, $x^{2} + y^{2} = 9$

$P A \times PB = P T^{2}$ [ By Geometry]
$P A \times PB = (3)^{2} + (11)^{2} - 9 = 121$