# 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...

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.

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,
$$y = \frac12 x+2$$