Question

If a circle whose centre is $(1,-3)$ touches the line $3 x-4 y-5=0$, then the radius of the circle is
(A) $2$
(B) $4$
(C) $\dfrac{5}{2}$
(D) $\dfrac{7}{2}$

Answer 1

Hint:We know that circle's tangent is always perpendicular to its radius. So we have to apply the general formula and substitute the given values to get the value of the relation. One of the crucial components of a circle is its radius. It is the separation between any point on a circle's edge and its centre.

Complete step by step Solution:
As we are aware, a circle's tangent is always perpendicular to its radius.
$R=\sqrt{\dfrac{Ax+By+C}{A{{{\hat{A}}}^{2}}+B{{{\hat{A}}}^{2}}}}$
Simplify the equation
$R=\dfrac{3\tilde{A}1+-4\tilde{A}(-3)-5}{\sqrt{3{{{\hat{A}}}^{2}}+(-4){{{\hat{A}}}^{2}}}}$
$R=\dfrac{3+12-5}{5}$
$R=\dfrac{10}{5}$
$R=2$
The radius of the circle is 2

Therefore, the correct option is A.

Additional information:
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 centre. When the radius's length varies, the circle's size also changes.

Note: Note that the group of points whose separation from a fixed point has a constant value is 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$\left(\mathrm{x}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}-\mathrm{y}_{1}\right)^{2}=\mathrm{r}^{2}$ whose centre is at $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$