Find the points located at a distance of 10 units from the point A (-5: 2) and the axis OX

Question

I'm stuggling with this one for like 2 days. I need help.

Answer 1

If you draw a circle of radius 10 with point A as the center point (all points on the circle are 10 units away from A). Then draw the lines y = 10 and y = -10 since these are all 10 units away from the x-axis (which is y=0). Your answer is the points where the lines and circle intersect.

You will see that y = -10 doesn't intersect with your circle but y = 10 does at two points.

You should be able to see what the intersection points are on the graph but you can also solve algebraically for the two points. Write the equation for your circle: (x+5)^2 + (y-2)^2 = 100.
Now you have a system of equations:
(x+5)^2 + (y-2)^2 = 100
y = 10

Plug 10 into the y for the equation of the circle and solve for x.

You end up with (x+5)^2 = 36
Take the square root of both sides but you have to solve for x+5 = 6 as well as x+5 = -6, since you can square 6 and -6 and get 36.

You end up with x = 1 and x = -11

So your points are (1, 10) and (-11, 10).

Hope that helps!