Question

Center of the circle $(x-3)^{2}+(y-4)^{2}=5$ is
A. $(3,4)$
B. $(-3,-4)$
C. $(4,3)$
D. $(-4,-3)$

Answer 1

Hint: Here we have to find the center of the circle. The center of a circle is a location inside the circle that is situated in the middle of the circumference. Initially we have to compare the given circle equation with the general equation of the circle and we can find the points.

Complete step by step solution:
The general equation of a circle is another name for the center of the circle formula. If the radius is r, the center's coordinates are $(h,k),$and any point on the circle is$(x, y)$, the center of the circle formula is as follows:
$(x-h)^{2}+(y-k)^{2}=r^{2}$
The equation for the center of the circle is another name for this. In the sections that follow, we'll use this formula to determine a circle's equation or center.
${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$
$(h,k)\to $ Center of circle,
$r$ is the radius of circle
The radius of a circle is the length of the straight line that connects the center to any point on its circumference. Because a circle's circumference can contain an endless number of points, a circle can have more than one radius.
$(h,k)=(3,4)$
Then the center of the circle$(x-3)^{2}+(y-4)^{2}=5$ is $(3,4)$

Option ‘A’ is correct

Note: We can say that the group of points whose separation from a fixed point has a constant value are represented by a circle. The radius of the circle, abbreviated $r$, is a constant that describes this fixed point, which is known as the circle's centre. The formula for a circle$\left({x}-{x}_{1}\right)^{2}+\left({y}-{y}_{1}\right)^{2}={r}^{2}$ whose centre is at $\left({x}_{1}, {y}_{1}\right)$