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Problem 690

# Two concentric circles have radii whose lengths are 2 in. and 6 in. Line $$\mathrm{m}$$ is drawn, in the accompanying figure, tangent to the smaller circle, (a) Describe fully the locus of points equidistant from the two circles, (b) Describe fully the locus of points at a given distance $$d$$ from line $$\mathrm{m}$$. (c) How many points are there which satisfy the conditions given in both parts (a) and (b) if: (1) $$\mathrm{d}<2$$ in.? (2) $$\mathrm{d}=2$$ in.? (3) $$\mathrm{d}=6$$ in.? (4) $$\mathrm{d}>6$$ in.?

Expert verified
In conclusion: - For $$d < 2$$, there are 2 points satisfying the conditions. - For $$d = 2$$, there is 1 point satisfying the conditions. - For $$d = 6$$, there is 1 point satisfying the conditions. - For $$d > 6$$, there are no points satisfying the conditions.
See the step by step solution

## Step 1: Equations of the concentric circles

Since the circles are concentric, they have the same center. Let's assume the center of both circles is at the origin (0, 0). Then the equations of the two circles are: $$C_1: x^2 + y^2 = (2)^2$$ (smaller circle) $$C_2: x^2 + y^2 = (6)^2$$ (larger circle)

## Step 2: Locus of points equidistant from the two circles

A point is equidistant from the two circles if its distance from both circles is the same. Let's call this point P(x, y). The distance from P to circle C1 is the absolute difference between the radius of C1 and the distance from P to the origin (center of C1). Similarly, the distance from P to circle C2 can be found as well. Now, let the distances be equal: $$|r_1 - \sqrt{(x^2 + y^2)}| = |r_2 - \sqrt{(x^2 + y^2)}|$$ Where $$r_1$$ is the radius of smaller circle (2 inches) and $$r_2$$ is the radius of the larger circle (6 inches). When we solve this equation, we find that the locus of points equidistant from the two circles is a circle with the same center as C1 and C2, and with a radius of 4 in. (the average of the radii of the two circles). The equation of this circle is: $$L_1: x^2 + y^2 = (4)^2$$

## Step 3: Locus of points at a given distance d from line m

Let's assume line m has the equation $$y = m_1x + b_1$$. To find the locus of points at a given distance d from line m, we'll use the distance formula. The equation of a line parallel to line m at a distance d is: $$y = m_1x + b_1 \pm d\sqrt{1 + m_1^2}$$ Let's call this new line L2. The equation of L2 depends on the value of d.

## Step 4: Intersection points between L1 and L2 for different values of d

To find the number of points that satisfy both conditions, we need to find the intersection points between L1 (locus of points equidistant from the two circles) and L2 (locus of points at a given distance d from line m). We do this for different values of d: (1) For $$d < 2$$: There will be 2 intersection points between L1 and L2. (2) For $$d = 2$$: There will be 1 intersection point (tangent) between L1 and L2. (3) For $$d = 6$$: There will be 1 intersection point (tangent) between L1 and L2. (4) For $$d > 6$$: There will be no intersection points between L1 and L2. In conclusion, For $$d < 2$$, there are 2 points satisfying the conditions. For $$d = 2$$, there is 1 point satisfying the conditions. For $$d = 6$$, there is 1 point satisfying the conditions. For $$d > 6$$, there are no points satisfying the conditions.

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