Orthogonal Circles

Geometry Level 5

( x − 153 ) 2 + y 2 = 18 5 2 ( x + 153 ) 2 + y 2 = 18 5 2 x 2 + ( y − 1380 ) 2 = 19 6 2 \large{\begin{aligned} (x-153)^2+y^2&=&185^2 \\ (x+153)^2+y^2&=&185^2 \\ x^2+(y-1380)^2&=&196^2 \\ \end{aligned} }

Above shows the equations for 3 circles.

Two circles are orthogonal to each other if where they intersect meets at right angles. The circle that is orthogonal to all three of the circles above can be written as:

x 2 + ( y − k ) 2 = r 2 . x^2+(y-k)^2=r^2.

Find k + r k+r .


The answer is 1352.

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Garrett Clarke by Garrett Clarke , 25, USA

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1 solution

Chew-Seong Cheong
Jul 13, 2015

Two circles x 2 + y 2 + 2 g x + 2 f y + c = 0 x^2 + y^2 + 2gx + 2fy + c = 0 and x 2 + y 2 + 2 g ′ x + 2 f ′ y + c ′ = 0 x^2 + y^2 + 2g'x + 2f'y + c' = 0 are orthogonal if 2 g g ′ + 2 f f ′ = c + c ′ 2gg' + 2ff' = c + c' (see Orthogonal Circles ).

Now we have:

{ ( x − 153 ) 2 + y 2 = 18 5 2 ⇒ x 2 + y 2 − 306 x − 10816 = 0 . . . ( 1 ) ( x + 153 ) 2 + y 2 = 18 5 2 ⇒ x 2 + y 2 + 306 x − 10816 = 0 . . . ( 2 ) x 2 + ( y − 1380 ) 2 = 19 6 2 ⇒ x 2 + y 2 − 2760 y + 1865984 = 0 . . . ( 3 ) x 2 + ( y − k ) 2 = r 2 ⇒ x 2 + y 2 − 2 k y + k 2 − r 2 = 0 . . . ( 4 ) \begin{cases} (x-153)^2 + y^2 = 185^2 & \Rightarrow x^2 + y^2 -306x -10816 = 0 &...(1) \\ (x+153)^2 + y^2 = 185^2 & \Rightarrow x^2 + y^2 +306x -10816 = 0 &...(2) \\ x^2 + (y-1380)^2 = 196^2 & \Rightarrow x^2 + y^2 -2760y + 1865984 = 0 &...(3) \\ x^2 + (y - k)^2 = r^2 & \Rightarrow x^2 + y^2 -2ky + k^2-r^2 = 0 &...(4) \end{cases}

Considering 2 g g ′ + 2 f f ′ = c + c ′ 2gg' + 2ff' = c + c' , we have

{ ( 1 ) & ( 4 ) : 0 = k 2 − r 2 − 10816 ⇒ k 2 − r 2 = 10816 . . . ( 5 ) ( 3 ) & ( 4 ) : 2760 k = k 2 − r 2 + 1865984 = 10816 + 1865984 ⇒ k = 680 ( 5 ) : k 2 − r 2 = 10816 ⇒ r 2 = 68 0 2 − 10816 ⇒ r = 672 \begin{cases} (1)\&(4): & 0 = k^2 - r^2 - 10816 \quad \Rightarrow \color{#3D99F6} {k^2 - r^2 = 10816} \quad ...(5) \\ (3)\&(4): & 2760k = \color{#3D99F6}{k^2 - r^2} + 1865984 = \color{#3D99F6}{10816} + 1865984 \quad \Rightarrow \color{#D61F06}{k = 680} \\ (5): & \color{#D61F06}{k^2} - r^2 = 10816 \quad \Rightarrow r^2 = \color{#D61F06}{680^2} - 10816 \quad \Rightarrow r = 672 \end{cases}

⇒ k + r = 680 + 672 = 1352 \Rightarrow k + r = 680 + 672 = \boxed{1352}

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Moderator note:

What is special about this center?

Hint: Power of a point.

I believe you meant that r = 672 r=672 , not r = 627 r=627 . Great solution though!

Garrett Clarke - 5 years, 11 months ago

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Thanks. I have changed it.

Chew-Seong Cheong - 5 years, 11 months ago

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