At what point in the first quadrant does the line with equation y = 2.5 x + 3 intersect the circle with radius 3 and center (0, 3)?

Question

At what point in the first quadrant does the line with equation y=2.5x+3 intersect the circle with radius 3 and center (0, 3)? 

Answer 1

I would not leave a radical in the denominator. Most math teachers will request that you rationalize denominators.

It's a simple process. 6/√29 = 6×√29 / (√29/√29) = 6√29 / 29.

Note that the radical is no longer in the denominator. And we did this by multiplying by 1, so the value of the number has not changed, only its appearance. √29 / √29 = 1.

Similarly, 15 / √29 should be rationalized as 15√29/29.

Answer 2

The circle with radius 3 and center (0, 3) has the equation x² + (y – 3)² = 9

To get the x-coordinate of our required point we replace y in this equation with 2.5x + 3 (from the equation for the intersecting line). If y = 2.5x + 3 then y – 3 = 2.5x, so we have

x² + (y – 3)² = x² + (2.5x)² = x² + 6.25x² = 7.25x² = 9

this means x² = 9/7.25 = 36/29 and x = √(36/29) = 6/√29

and y = 2.5x + 3 = 2.5(6/√29) + 3 = 15/√29 + 3

answer

point in the first quadrant where the line intersects the circle is (6/√29,15/√29 + 3)

checking our answer

x² + (y – 3)² = (6/√29)² + (15/√29)² = 36/29 +225/29 = 261/29 = 9

Note

A circle with radius r and center (a,b) has the equation (x – a)² + (y – b)² = r²