A tangent is drawn at any point P on the parabola $y^2=8x$ and on is taken a point $Q(\alpha, \beta)$ from which tangents QA and QB are drawn to the circle $x^2+y^2=4$ Find the locus of the circumcentre of $\Delta AQB$ If P(8,8)
The equation of tangent the circle is $$x-2y+8=0$$$(\alpha,\beta)$ lies on this line
The tangents are drawn from this point to the circle, but no data has been provided regarding their slopes etc .
The tangents to the circle will be of the form
$$\beta =m\alpha \pm 2\sqrt {1+m^2}$$
There really isn’t much info to salvage, so that’s all I could go. I think there is equation for the circumcentre of a triangle, but I can’t recall it.
Best Answer
Note that QA $\perp$ OA and QB $\perp$ OB. Therefore, QAOB is cyclic with OQ being the diameter of the circumcircle. Since Q is on the line $x-2y+8=0$, let Q$(a, \frac a2+4)$. Then, the circumcenter is halfway on the diameter line OQ and it coordinates are,
$$x = \frac a2,\>\>\>\>\> y = \frac a4+2$$
Eliminate $a$ to obtain its locus,
$$y = \frac12 x+2$$