Radius of a Circle: Solving Negative Equations

chemic_23 · Dec 15, 2008

Homework Statement

Left: a coordinate diagram with two circles tangent to the lines x - 2y + 2 = 0 and 2x - y - 17 = 0, labelled C(2, -3) and C(58/9, 13/9), with point (6, -1) marked. Right: a worked solution with tangency equations (1) and (2) using distance formulas, and equation (3) for the point (6, -1) on the circle.

Homework Equations

Why is there negative on h-2k+2?and a denominator of negative square root of 5?
also in 2h-k-17, it has a negative sign but the denominator is not negative square root of 5? I really don't understand this part... please explain

Two equations for the radius r of a circle, each giving r as a distance formula. Equation (1): (h - 2k + 2) over negative square root of 5 equals r, with note (x - 2y + 2 = 0 is tangent to the circle). Equation (2): (2h - k - 17) over square root of 5 equals r, with note (2x - y - 17 = 0 is tangent to the circle).

The Attempt at a Solution

rl.bhat · Dec 15, 2008

In this problem center of the circle lies between the two tangents. In such cases, to find the perpendicular distances, you have to make the constant C in the tangent positive.

Radius of a Circle: Solving Negative Equations

Homework Statement

Homework Equations

The Attempt at a Solution

Discussion

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