A circle is given by x^2 + y^2 - 6x + 8y - 11 | vedclass

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

Question

(B) Let the equation of the circle be $S(x, y) = x^2 + y^2 - 6x + 8y - 11 = 0$. For point $P(0, 0)$: $S(0, 0) = 0^2 + 0^2 - 6(0) + 8(0) - 11 = -11$. S

Vedclass · Accepted Answer

One outside and one inside the circle

Vedclass · Answer

(B) Let the equation of the circle be $S(x, y) = x^2 + y^2 - 6x + 8y - 11 = 0$. For point $P(0, 0)$: $S(0, 0) = 0^2 + 0^2 - 6(0) + 8(0) - 11 = -11$. Since $S(0, 0) For point $Q(1, 8)$: $S(1, 8) = 1^2 + 8^2 - 6(1) + 8(8) - 11 = 1 + 64 - 6 + 64 - 11 = 112$. Since $S(1, 8) > 0$,the point $Q(1, 8)$ lies outside the circle. Therefore,one point lies inside and one point lies outside the circle.

$A$ circle is given by $x^2 + y^2 - 6x + 8y - 11 = 0$. Determine the position of the points $P(0, 0)$ and $Q(1, 8)$ with respect to the circle.

Similar Questions

The equation of the circle passing through the points $(1, 0)$ and $(0, 1)$ with the smallest radius is . . . . . .

If the endpoints of a diameter of a circle are $(1, 2)$ and $(3, 4)$,then its equation is:

The equation of a circle of radius $5$ units touching another circle $x^2+y^2-2x-4y-20=0$ at $(5,5)$ is

If $(\alpha, \beta)$ is the centre of a circle passing through the origin,then its equation is

The equation of a circle touching the coordinate axes and the line $x \cos \alpha + y \sin \alpha = 2$ can be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module