What is the ratio of areas of circle C1 & circle C2?

Question

What is the ratio of areas of circle C1 with center at (1,-4) & circle C2 with center (5,-8) both having a common tangent line L with equation y = x +3?

A. 1:2
B. 4:1
C. 1:4
D. 2:1
E. 1:1

firas92 · Accepted Answer

Perpendicular distance from a point  (m,n)  to a line  Ax+By+C=0  is given by

D=\frac{|Am+Bn+C|}{\sqrt{A^2+B^2}}

Rearranging the equation as  x-y+3=0  and using this formula, we can calculate the radius of the two circles (since by property, the tangent to a circle is perpendicular to its radius at the point of intersection)

We get,  R1 = 4\sqrt{2}  and  R2=8\sqrt{2}

Therefore Areas,  A1=32π  and  A2=128π

So, ratio of areas =  \frac{1}{4}

Answer is  (C)

warrior1991 · Answer

chetan2u  Is there any other way than solving this question via a formula of a distance??

warrior1991 · Answer

generis

Is there any way other than using a formula to solve the problem??

IanStewart · Answer

You can do the problem without a special formula. A tangent line to a circle is perpendicular to the circle's radius at the tangent point. So if we take the first circle, with center (1, -4), and draw its radius to the tangent line, the slope of that radius will be -1 (since it is perpendicular to y = x + 3). If the slope is -1, the equation of that radius-line is y = -x + b, and since the center (1, -4) is on that line, we can substitute into the line's equation to find b: 1 = 4 + b, so b = -3. So the equation of the radius-line is y = -x - 3. If we solve for where this radius meets the tangent line, by solving the two equations for the lines together, we find they meet at (-3, 0). From (1, -4) to (-3, 0) we go across 4 units and up 4 units, so using Pythagoras, if r is the radius, r^2 = 4^2 + 4^2 = 32, and the area of the circle is 32π.

You could do the same for the second circle too, but if you notice the center of that circle, (5, -8), is also on the radius-line we just found, you can see immediately that it will also meet the tangent line at the same point, (-3, 0). Since we go across 8 units and up 8 units from (5, -8) to (-3, 0), then by Pythagoras, if R is the radius, R^2 = 8^2 + 8^2 = 128, and the area of the larger circle is 128π.

So the ratio of the two areas is 32 to 128, or 1 to 4.

You could do the same for the second circle too, but if you notice the center of that circle, (5, -8), is also on the radius-line we just found, you can see immediately that it will also meet the tangent line at the same point, (-3, 0). Since we go across 8 units and up 8 units from (5, -8) to (-3, 0), then by Pythagoras, if R is the radius, R^2 = 8^2 + 8^2 = 128, and the area of the larger circle is 128π.

So the ratio of the two areas is 32 to 128, or 1 to 4.

pratik2018 · Answer

Plotting the points on the graph shows that C1 < C2 so only options remaining are A and C - C2 seems to be twice as far away from the line as compared to the C1. radius almost equal to twice as C1. For calculating area, squaring the radius makes it four times. C is the correct option.

philipssonicare · Answer

What is the question source?

bumpbot · Answer

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What is the ratio of areas of circle C1 & circle C2?

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What is the ratio of areas of circle C1 & circle C2?

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