The number of lines that can be drawn through the point (5, 2) at a distance of 5 units from the point (2, -2) is

Let A (5,  2) and m be the slope of line passing through A

Equation of line

$\begin{array}{l}y-2=m\left(x-5\right)--\left(1\right)\\ \Rightarrow mx-y+2-5m=0\\ Given\perp rdis\tan cefrom\left(2,-2\right)to\left(1\right)=5\\ \Rightarrow \frac{\left|2m+2+2-5m\right|}{\sqrt{m^2+1}}=5\\ \Rightarrow \left|4-3m\right|=5\sqrt{m^2+1}\\ S.B.S\\ \Rightarrow 16+9m^2-24m=25m^2+25\\ \Rightarrow 16m^2+24m+9=0\\ \Delta =b^2-4ac={\left(24\right)}^2-4\left(16\right)9\Rightarrow 576-576\Rightarrow 0\end{array}$

m  has only one value

No of lines are “ one “