A circle pass through origin and centre is $(3,-3)$ and find coordinated point on the circle

Question

A circle pass through origin and centre is $(3,-3)$ and line $y=x-6$ meet the circle at point $P$ and $Q$. Find coordinated of point on the circle where tangent are parallel to line $PQ$. I got the equation $(x-3)^2+(y+3)^2=18$
Coordinate $P(6,0)$, $Q(0,-6)$.
I don't know how to find coordinated point!! Anyone can help me?

Answer 1

Let the equation of the circle be $$(x-3)^2+(y+3)^2=r^2\ \ \ \ (1)$$

As the circle pass through origin, $r^2=(0-3)^2+(0+3)^2=18$

$$\implies x^2+y^2-6x+6y=0\ \ \ \ (2)$$

Now the equation of line parallel to $y=x-6$ will be $y=x+c$ where $c$ is an arbitrary constant

Replacing $y$ in $(2),$

$$x^2+cx+3c=0$$ each root of represents the abscissa of intersection.

For tangency, both roots must be same $$\implies0=c^2-4(3c)=c(c-12)$$

$c=0\implies x=0$ and $c=12\implies x=?$

Answer 2

Notice, the equation of the circle with center at $(3, -3)$ & passing through the origin $(0, 0)$ is given as $$(x-3)^2+(y+3)^3=\left(\sqrt{(0-3)^2+(0+3)^2}\right)^2=18$$ $$x^2+y^2-6x+6y=0$$ differentiating w.r.t. $x$, $$2x+2y\frac{dy}{dx}-6+6\frac{dy}{dx}=0$$ $$\frac{dy}{dx}=\frac{3-x}{3+y}$$

Now, substituting $y=x-6$ in the equation of the circle $$x^2+(x-6)^2-6x+6(x-6)=0$$ $$x=6, 0\implies y=0, -6$$ hence, the points of intersection $P(6, 0)$ & $Q(0, -6)$ since, the tangent to the circle is parallel to the line $PQ$ hence $$\frac{dy}{dx}=\text{slope of line PQ}$$ $$\frac{3-x}{3+y}=\frac{0-(-6)}{6-0}=1$$ $$y=-x\tag 1$$ substituting $y=-x$ in the equation of the circle, one should get $$x^2+(-x)^2-6x+6(-x)=0$$ $$x(x-6)=0$$$$x=6, 0\implies y=0, -6$$ hence, there are two points, on the circle: $x^2+y^2-6x+6y=0$, $\color{red}{(0, 0)}$ & $\color{red}{(6, -6)}$ at which tangents drawn are parallel to the given line: $y=x-6$.