In the xy-plane,the segment with endpoints (3 | vedclass

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

Question

(B) The center of the circle is the midpoint of the diameter: $(\frac{3-5}{2}, \frac{8+2}{2}) = (-1, 5)$.The radius squared $r^2$ is the distance from

Vedclass · Accepted Answer

exactly one integral $k$

Vedclass · Answer

(B) The center of the circle is the midpoint of the diameter: $(\frac{3-5}{2}, \frac{8+2}{2}) = (-1, 5)$.The radius squared $r^2$ is the distance from the center $(-1, 5)$ to $(3, 8)$ squared: $r^2 = (3 - (-1))^2 + (8 - 5)^2 = 4^2 + 3^2 = 16 + 9 = 25$.The equation of the circle is $(x + 1)^2 + (y - 5)^2 = 25$.Substitute the point $(k, 10)$ into the equation: $(k + 1)^2 + (10 - 5)^2 = 25$.$(k + 1)^2 + 5^2 = 25$.$(k + 1)^2 + 25 = 25$.$(k + 1)^2 = 0$.$k + 1 = 0$,so $k = -1$.Since $k = -1$ is an integer,there is exactly one integral value of $k$.

In the $xy$-plane,the segment with endpoints $(3, 8)$ and $(-5, 2)$ is the diameter of a circle. The point $(k, 10)$ lies on the circle for:

Similar Questions

The equation of the circle with the origin as the centre passing through the vertices of an equilateral triangle whose median is of length $3a$ is

$A$ circle is concentric with the circle $x^2 + y^2 - 6x + 12y + 15 = 0$ and has an area double that of the given circle. The equation of the circle is:

The equation of a circle with center at $(-2, 3)$ and circumference of $4 \pi$ units is

The equation of the circle whose centre lies on the lines $3x - y - 4 = 0$ and $x + 3y + 2 = 0$ and has an area of $154$ square units is:

Find the radius and center of the circle $2x^2 + 2y^2 = 3x - 5y + 7$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module