Does the point (-2.5, 3.5) lie inside,outside | vedclass

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

Question

(A) The equation of the given circle is $x^{2}+y^{2}=25$. This is of the form $(x-h)^{2}+(y-k)^{2}=r^{2}$,where the centre $(h, k) = (0, 0)$ and the r

Vedclass · Accepted Answer

Inside the circle

Vedclass · Answer

(A) The equation of the given circle is $x^{2}+y^{2}=25$. This is of the form $(x-h)^{2}+(y-k)^{2}=r^{2}$,where the centre $(h, k) = (0, 0)$ and the radius $r = \sqrt{25} = 5$. To determine the position of the point $P(-2.5, 3.5)$ relative to the circle,we calculate the distance $d$ from the centre $(0, 0)$ to the point $P$: $d = \sqrt{(-2.5 - 0)^{2} + (3.5 - 0)^{2}}$ $d = \sqrt{6.25 + 12.25}$ $d = \sqrt{18.5} \approx 4.3$. Since $d \approx 4.3 Therefore,the point $(-2.5, 3.5)$ lies inside the circle.

Does the point $(-2.5, 3.5)$ lie inside,outside,or on the circle $x^{2}+y^{2}=25$?

Similar Questions

Let $S$ be the circumcircle of the triangle formed by the line $x-2y-4=0$ with the coordinate axes. If $P(-2, -4)$ is a point in the plane of the circle $S$ and $Q$ is a point on $S$ such that the distance between $P$ and $Q$ is the least,then $PQ=$

The centre of the circle that passes through the point $(0,1)$ and touches the curve $y=x^2$ at $(2,4)$ is

Find the equation of the circumcircle of the triangle formed by the lines $y \pm \sqrt{3}x = 6$ and the $x$-axis.

Find the equation of the chord of the circle $x^2 + y^2 - 6x + 8y = 0$ which is bisected at the point $(5, -3)$.

Consider the circles $x^2 + (y - 1)^2 = 9$ and $(x - 1)^2 + y^2 = 25$. Which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module