Question

Prove that point (1,3) lies outside of circle $x^2+y^2-10x-y+4=0$

Answer 1

$x^2+y^2-10x-y+x=0$

$x^2-10x+25+y^2-y+\frac{1}{4}-\frac{1}{4}+4-25=0$

$(x-5{)}^2+{(y-\frac{1}{2})}^2=21+\frac{1}{4}=\frac{85}{4}$

$(x-5{)}^2+{(y-\frac{1}{2})}^2={\left(\frac{\sqrt{85}}{2}\right)}^2$

Centre$=(9,\frac{1}{2})$ and radius $\frac{\sqrt{85}}{2}$

If distance of Pt gram centre is less than radius then pt lies inside circle.

$(1,3)$, Distance between centre card $(1,3)$

$=\sqrt{(5-1{)}^2+{(\frac{1}{2}-3)}^2}=\sqrt{16+\frac{25}{4}}=\frac{1}{2}\sqrt{89}$

$\frac{\sqrt{89}}{2}>\frac{\sqrt{85}}{2}$, so Pt lies outside the circle.