$A$ line drawn through the point $P(4, 7)$ cuts the circle $x^2 + y^2 = 9$ at the points $A$ and $B$. Then $PA \cdot PB$ is equal to

  • A
    $53$
  • B
    $56$
  • C
    $74$
  • D
    $65$

Explore More

Similar Questions

Given the circle $C$ with the equation $x^2+y^2-2x+10y-38=0$. Match the List-$I$ with the List-$II$ given below concerning $C$.
List-$I$List-$II$
$A$. The equation of the polar of $(4, 3)$ with respect to $C$$I$. $y+5=0$
$B$. The equation of the tangent at $(9, -5)$ on $C$$II$. $x=1$
$C$. The equation of the normal at $(-7, -5)$ on $C$$III$. $3x+8y=27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$$IV$. $x=9$

Let $C_i \equiv x^2 + y^2 = i^2$ for $i = 1, 2, 3$ be three circles. There are $4i$ points on the circumference of each circle $C_i$. If no three of all the points on the three circles are collinear,then the number of triangles that can be formed using these points whose circumcentre does not lie on the origin is:

If the equation of the circumcircle of the triangle formed by the lines $L_1 \equiv x+y=0$,$L_2 \equiv 2x+y-1=0$,and $L_3 \equiv x-3y+2=0$ is $\lambda_1 L_1 L_2 + \lambda_2 L_2 L_3 + \lambda_3 L_3 L_1 = 0$,then find the value of $\frac{7 \lambda_1}{\lambda_2} + \frac{\lambda_3}{\lambda_1}$.

$A$ circle is inscribed in an equilateral triangle of side $a$. The area of any square inscribed in the circle is

$A$ focal chord to $y^2 = 16x$ is a tangent to $(x - 6)^2 + y^2 = 2$. Then the possible values of the slope of this chord are:

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo