Answer:
[tex]\large \boxed{(-1,-6),(-1,10)}[/tex]
Step-by-step explanation:
Point A must be on a circle with centre at (5,2) and radius 10.
The centre-radius form for the equation of a circle is
(x – h)² + (y – k)² = r², where
(h,k) = the coordinates of the centre and
r = the radius
Thus, the equation for this circle is
(x - 5)² + ( y - 2)² = 10²
Substitute the value for x and solve for y:
[tex]\begin{array}{rcl}\\(x - 5)^{2} +( y - 2)^{2} &=& 10^{2}\\(-1 - 5)^{2} +( y - 2)^{2} &=& 100\\(-6)^{2} +( y - 2)^{2} &=& 100\\36 +( y - 2)^{2} &=& 100\\( y - 2)^{2} &=& 64\\y - 2 = -8 & \qquad & y - 2 = 8\\y = \mathbf{-6} & \qquad & y = \mathbf{10}\\\end{array}\\\text{The possible coordinates of A are $\large \boxed{\mathbf{(-1,-6),(-1,10)}}$}[/tex]
The figure below shows the circle intersecting the line x = -1 at (-1,-6) and (-1,10).
