The least and the greatest distances of the p | vedclass

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

Question

(A) Given the circle equation: $x^{2} + y^{2} - 4x - 2y - 20 = 0$. Comparing with $x^{2} + y^{2} + 2gx + 2fy + c = 0$,we get $g = -2, f = -1, c = -20$

Vedclass · Accepted Answer

$5, 15$

Vedclass · Answer

(A) Given the circle equation: $x^{2} + y^{2} - 4x - 2y - 20 = 0$. Comparing with $x^{2} + y^{2} + 2gx + 2fy + c = 0$,we get $g = -2, f = -1, c = -20$. Centre of the circle is $(-g, -f) = (2, 1)$. Radius $r = \sqrt{g^{2} + f^{2} - c} = \sqrt{(-2)^{2} + (-1)^{2} - (-20)} = \sqrt{4 + 1 + 20} = \sqrt{25} = 5$. Let $P$ be the point $(10, 7)$ and $C$ be the centre $(2, 1)$. The distance $d$ between $P$ and $C$ is $\sqrt{(10 - 2)^{2} + (7 - 1)^{2}} = \sqrt{8^{2} + 6^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10$. The least distance of the point from the circle is $d - r = 10 - 5 = 5$. The greatest distance of the point from the circle is $d + r = 10 + 5 = 15$. Thus,the least and greatest distances are $5$ and $15$ respectively.

The least and the greatest distances of the point $(10, 7)$ from the circle $x^{2} + y^{2} - 4x - 2y - 20 = 0$ are

Similar Questions

The number of circles that touch all the three lines $x+y-1=0$,$x-y-1=0$,and $y+1=0$ is

Find the equation of the director circle of the circle $x^2 + y^2 = 8$.

Let $ABCD$ be a quadrilateral with area $18$,with side $AB$ parallel to the side $CD$ and $AB = 2CD$. Let $AD$ be perpendicular to $AB$ and $CD$. If a circle is drawn inside the quadrilateral $ABCD$ touching all the sides,then its radius is

If a circle $C$ passing through the point $(4, 0)$ touches the circle $x^2+y^2+4x-6y=12$ externally at the point $(1, -1)$,then the radius of $C$ is

The line $4x + 3y - 4 = 0$ divides the circumference of a circle in the ratio $1:2$. If $C(5, 3)$ is the center of that circle,then the equation of the circle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module