Looks are deceptive

Geometry Level 3

The circle $\alpha$ has center at $(1,2)$ and radius $3$ . The circle $\beta$ has center at $(9,8)$ and radius $7$ . They touch at the point $(a,b)$ where $a$ and $b$ are fractions with same denominator. Find the sum of their numerators and denominators (taken only once).

Note

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69

74

46

41

1 solution

Brian Charlesworth
Nov 5, 2014

The point $(a,b)$ will lie on the line joining the centers of the two circles. This line has slope $\frac{8 - 2}{9 - 1} = \frac{3}{4}$ , and since it passes through $(1,2)$ will have the equation

$y - 2 = (\frac{3}{4})(x - 1)$ . Thus $b - 2 = (\frac{3}{4})(a - 1)$ , (i).

Now $(a,b)$ must also satisfy the equation $(x - 1)^{2} + (y - 2)^{2} = 9$ , and so we have that $(a - 1)^{2} + (b - 2)^{2} = 9$ . Now substitute (i) into this equation to find that

$(a - 1)^{2} + (\frac{9}{16})(a - 1)^{2} = 9 \Longrightarrow (\frac{25}{16})(a - 1)^{2} = 9 \Longrightarrow a - 1 = \frac{12}{5} \Longrightarrow a = \frac{17}{5}$ .

Then from (i) we have that $b = 2 + (\frac{3}{4})(\frac{12}{5}) = \frac{19}{5}$ .

Thus the desired sum is $17 + 19 + 5 = \boxed{41}$ .

Section formula would be much easier

Krishna Sharma - 6 years, 7 months ago

True. :)

$(\frac{3}{10})\langle 8, 6 \rangle + \langle 1, 2 \rangle = \langle \frac{17}{5}, \frac{19}{5} \rangle$ .

Sometimes I just start typing without thinking first. It's a really bad habit. :(

Brian Charlesworth - 6 years, 7 months ago

I also did the same.

Shyambhu Mukherjee - 5 years, 7 months ago

Looks are deceptive

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