Question

A circle passes through the points $(-1,1)$ and has center $(2,0)$. Find the points on the circle, tangent at which are parallel to the straight line joining origin to the center

Answer 1

The center of circle is Given as $(2,0)$

The equation of circle is given as $(x-2{)}^2+(y-0{)}^2=r^2$

The point it passes through is $(-1,4)$

$(-1-2{)}^2+4^2=r^2{r}^2=9+16r=\sqrt{25}r=5$

So the equation of circle is $x^2+y^2-4x+4=25x^2+y^2-4x-21=0$

Slope of line passing through origin and center $\frac{0-0}{2-0}=0$

Slope of tangent is given as $2x+2y\frac{dy}{dx}-4=0\frac{dy}{dx}=\frac{2-x}{y}=0x=2$

So when $x=2⟹y=\pm 5$

So tangent at $(2,5)$ is $y-5=0(x-2)⟹y=5$

So tangent at $(2,-5)$ is $y+5=0(x-2)⟹y=-5$