Question

If the radius of the circle $x^{2}+y^{2}-18 x-12 y+k=0$ be 11 then $k=$
A) 34
B) 4
C) -4
D) 49

Answer 1

Hint: A circle is a two-dimensional object made up of points that are spaced out from a given point (center) on the plane by a fixed or constant distance (radius). The fixed point is referred to as the circle's origin or center, and the fixed distance between each point and the origin is referred to as the radius.

Complete Step by step solution:
The equation of the circle is $x^{2}+y^{2}-18 x-12 y+k=0$.
Compare it with $\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0$ we get,
$2 \mathrm{~g}=-18,2 \mathrm{f}=-12$ and $\mathrm{c}=\mathrm{k}$
$\therefore \mathrm{g}=-9, \mathrm{f}=-6$ and $\mathrm{c}=\mathrm{k}$
Center $=(-\mathrm{g},-\mathrm{f})=(9,6)$
The radius of a circle is the length of the straight line that connects the centre 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. This indicates that a circle has an endless number of radii and that each radius is equally spaced from the circle's center. When the radius's length varies, the circle's size also changes.
Radius of a circle is given $(r)=11 \quad$
We know,
$\Rightarrow \mathrm{r}=\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}}$
$\Rightarrow 11=\sqrt{(-9)^{2}+(-6)^{2}-\mathrm{k}}$
$\Rightarrow 11=\sqrt{81+36-\mathrm{k}}$
$\Rightarrow 11=\sqrt{117-\mathrm{k}}$
$\Rightarrow 121=117-\mathrm{k}$
$\Rightarrow 4=-\mathrm{k}$
$\therefore \mathrm{k}=-4$
So the correct answer is option(C).

Note: The equation for a circle has the generic form: ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$. The coordinates of the circle's center and radius are found using this general form, where g, f, and c are constants. The general form of the equation of a circle makes it difficult to identify any significant properties about any specific circle, in contrast to the standard form, which is simpler to comprehend. So, to quickly change from the generic form to the standard form, we will use the completing the square formula.