The point P(10, 7) lies outside the circle x^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given equation of the circle is $x^2 + y^2 - 4x - 2y - 20 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(D) The given equation of the circle is $x^2 + y^2 - 4x - 2y - 20 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -1$,and $c = -20$.The center of the circle is $C(-g, -f) = (2, 1)$.The radius of the circle is $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (-1)^2 - (-20)} = \sqrt{4 + 1 + 20} = \sqrt{25} = 5$.The distance between point $P(10, 7)$ and the center $C(2, 1)$ is $CP = \sqrt{(10 - 2)^2 + (7 - 1)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$.The greatest distance of point $P$ from the circle is given by $CP + r = 10 + 5 = 15$.

The point $P(10, 7)$ lies outside the circle $x^2 + y^2 - 4x - 2y - 20 = 0$. The greatest distance of $P$ from the circle is

Similar Questions

The tangents are drawn from the point $(4, 5)$ to the circle $x^2 + y^2 - 4x - 2y - 11 = 0$. The area of the quadrilateral formed by these tangents and the radii is .............. $sq. \text{ units}$.

The area of a circle whose centre is $(h, k)$ and radius $a$ is

$A$ line drawn through the point $A(5,7)$ cuts the circle $x^2+y^2-36=0$ at the points $P$ and $Q$. Then,$AP \cdot AQ=$

The radius of any circle touching the lines $3x - 4y + 5 = 0$ and $6x - 8y - 9 = 0$ is

The line $y = x$ cuts the circle $x^2 + y^2 - 2x = 0$ at points $A$ and $B$. Find the equation of the circle having $AB$ as its diameter.

Vedclass Test Series

Exam Paper Generator

Online Exam Module