Orthogonal Circles

Relevant wiki: Power of a Point

By the power of a point , we have

$(OP)^2=(OB)(OD)$

but $OB=10-OP$ (since $AO=OP$ ) and $OD=OP+24$ (since $OC=OP$ ), therefore

$(OP)^2=(10-OP)(OP+24)$

$(OP)^2=10(OP)+240-(OP)^2-24(OP)$

$2(OP)^2+14(OP)-240=0$

$(OP)^2+7(OP)-120=0$

$(OP+15)(OP-8)=0$

$OP=-15$ or $OP=8$ (reject the negative value)

So $OP=8$

It follows that $BC=2(8)-10=$ $\boxed{6}$

Alternate Solution: let $d$ and $r$ be the diameter and radius of the small circle respectively; let $D$ and $R$ be the diameter and radius of the big circle respectively; let $x$ be the length of the red line

From the figure, $d=10+x$ , therefore $r=5+\dfrac{x}{2}$ and $D=24+x$ , therefore $R=12+\dfrac{x}{2}$

$OB=10-r=10-\left(5+\dfrac{x}{2}\right)=5-\dfrac{x}{2}$

$QC=24-R=24-\left(12+\dfrac{x}{2}\right)=12-\dfrac{x}{2}$

The length of the hypotenuse of the right triangle is therefore, $OQ=5-\dfrac{x}{2}+x+12-\dfrac{x}{2}=17$

Applying pythagorean theorem in the right triangle, we have

$17^2=r^2+R^2$ $\implies$ $289=\left(5+\dfrac{x}{2}\right)^2+\left(12+\dfrac{x}{2}\right)^2$ $\implies$ $289=25+5x+\dfrac{x^2}{4}+144+12x+\dfrac{x^2}{4}$

After simplifying, the result is a quadratic equation

$x^2+34x-240$

solving for $x$ using the quadratic formula gives

$x=$ $\boxed{6}$

From the figure,

$2x = 10 + m$ ; $x = 5 + \frac{m}{2}$

$2y = 24 + m$ ; $y = 12 + \frac{m}{2}$

$z = 10 + m + 24 - (5 + \frac{m}{2}) - (12 + \frac{m}{2}) = 17$

By pythagorean theorem,

$x^2 + y^2 = z^2$

$(5 + \frac{m}{2})^2 + (12 + \frac{m}{2})^2 = 17^2$

$m^2 + 34m - 240 = 0$

Solving for $m$ ,

$m = BC = 6$

Relevant wiki: Power of a Point

Let radius $OP = r$ , $PQ = R$ , and $OQ = s$ . By Pythagorean theorem, $s^2 = r^2 + R^2$ .

The desired $BC = R + r - s$ and $AD = R + r +s$ .

Therefore, $BD\cdot AD = ((R+r)-s)((R+r)+s) = (R+r)^2 - s^2 = 2Rr$

Then $AB = 10 = 2r - (R+r-s) = r-R+s$ , and $CD = 24 = 2R - (R+r-s) = R-r+s$ .

Hence, $AB\cdot CD = (s - (R-r))(s + (R-r)) = s^2 - (R-r)^2 = 2Rr$ .

That means $AD\cdot BC = AB\cdot CD = 240$ .

Suppose $BC = x$ .

Then $x(x+10+24) = 240$ .

$x^2 + 34 -240 = 0$

$(x-6)(x+40) = 0$

Thus, $x=\boxed{6}$ .

For the radii, $r = \dfrac{10+6}{2} = 8$ ; $R = \dfrac{24+6}{2} = 15$ .

Finally, $s = \sqrt{8^2 + 15^2} = 17$ .

Note : The ratios along points $A,B,C,D$ are called Harmonic range .