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Write an equation for a circle centered at (2,0) and tangent to y=x+2

Write an equation for a circle centered at (2,0) and tangent to y=x+2-example-1
User Tim Fuqua
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Answer:

(a) (x -2)² +y² = 8

Explanation:

The distance between the center and the point of tangency at (0, 2) is ...

d = √((x2 -x1)² +(y2 -y1)²)

d = √((0 -2)² +(2 -0)²) = √8

This is the radius of the circle. The circle centered at (h, k) with radius r has equation ...

(x -h)² +(y -k)² = r²

For the given values (h, k) = (2, 0) and r = √8, the equation of the circle is ...

(x -2)² +y² = 8

_____

Additional comment

The tangent line has a slope of 1. The radius to the point of tangency will be perpendicular, so will have a slope of -1. The point of tangency will be where the lines y=x+2 and y=-(x-2) meet, at the point (0, 2).

Above, we found the radius using the distance formula. You can also find it using the "distance to a line" formula:

d = |ax +by +c|/√(a²+b²)

where d is the distance from (x, y) to line ax+by+c = 0. Our given line is x-y+2=0, so the distance from point (2, 0) to the tangent line is ...

d = |1(2)-1(0)+2|/√(1²+(-1)²) = 4/√2 = √8

Write an equation for a circle centered at (2,0) and tangent to y=x+2-example-1
User Eric Chu
by
7.5k points

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