
Equation of a circle with centre $(4,3)$ touching the circle $x^{2}+y^{2}=1$ is
A) $\quad x^{2}+y^{2}-8 x-6 y+9=0$
B) $\quad x^{2}+y^{2}+8 x+6 y-11=0$
C) $\quad x^{2}+y^{2}-8 x-6 y-11=0$
D) $x^{2}+y^{2}+8 x+6 y-9=0$
Answer
164.7k+ views
Hint: 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 square formula.
Complete step by step Solution:
Let equation of circle be ${{(x-4)}^{2}}+{{(y-3)}^{2}}={{r}^{2}}..............(1)$
If the circle (1) touches the circle $x^{2}+y^{2}=1$,
then the distance between the centers $(4,3)$ and $(0,0)$ of these circles is equal to the sum or difference of their radii, $r$ and 1
$\Rightarrow \sqrt{\left(4^{2}-0\right)+\left(3^{2}-0\right)}=1+r$
$\Rightarrow r+1=5$
$\Rightarrow r=4$
Consequently, the necessary circles' equations from (1) are
$(x-4)^{2}+(y-3)^{2}=r^{2}$
$x^{2}-8 x+16+y^{2}-6 y+9=16$
$x^{2}+y^{2}-8 x-6 y+9=0$
Hence, the correct option is (D).
Additional Information: 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. 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.
Note: A circle's centre and radius make up its 2D shape. If we are aware of the circle's center and radius, we can draw any circle. The radii of a circle are infinitely variable. The midpoint where all of the radii meet is the center of a circle. The center of the circle's diameter is another way to describe it.
Complete step by step Solution:
Let equation of circle be ${{(x-4)}^{2}}+{{(y-3)}^{2}}={{r}^{2}}..............(1)$
If the circle (1) touches the circle $x^{2}+y^{2}=1$,
then the distance between the centers $(4,3)$ and $(0,0)$ of these circles is equal to the sum or difference of their radii, $r$ and 1
$\Rightarrow \sqrt{\left(4^{2}-0\right)+\left(3^{2}-0\right)}=1+r$
$\Rightarrow r+1=5$
$\Rightarrow r=4$
Consequently, the necessary circles' equations from (1) are
$(x-4)^{2}+(y-3)^{2}=r^{2}$
$x^{2}-8 x+16+y^{2}-6 y+9=16$
$x^{2}+y^{2}-8 x-6 y+9=0$
Hence, the correct option is (D).
Additional Information: 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. 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.
Note: A circle's centre and radius make up its 2D shape. If we are aware of the circle's center and radius, we can draw any circle. The radii of a circle are infinitely variable. The midpoint where all of the radii meet is the center of a circle. The center of the circle's diameter is another way to describe it.
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