2 points, A and B, on x-axis are intersections of x-axis and the circle with radius 2 unit drawn around a center P(-1,1). Point A is on the left of P, and B is on the right.
1. A solution by algebra
An equation of circle, where the radius is r, and the coordinates of center is (a,b) is: (x-a)²+(x-b)²=r²
Plug r=2, a=-1, and b=1 into the equation shown above. We have:
(x+1)²+(y-1)²=2² ··· Eq.1 The equation of x-axis is y=0. Plug y=0 lnto Eq.1 (x+1)²+(0-1)²=2²
We have: x²+2x-2=0 ··· Eq.2 Use the quadratic formula. We have: x=-1-√3, and x=√3-1
Therefore,the 2 points are A(-1-√3, 0), and B(√3-1, 0)
2. Solution by geometry
Let the foot of a perpendicular from P(-1,1) to x-axis be R(-1,0).
∠PRA=∠PRB=90°, AP=BP=2, and PR is mutual, so ΔAPR≡ΔBPR (RHS),and AR=BR
Here, AP=BP=2 and PR=1, so that 2 triangles are 1-2-√3 right triangles, and AR=BR=√3
Threrefore, the 2 points are A(-1-√3, 0), and B(√3-1, 0).