3 Circles and a Line

$\begin{aligned} \sin(\alpha) &= \frac{DF}{OF}\\ &=\frac15 \\ \frac{OG}{\sin(\beta)}&=\frac{AG}{\sin(\alpha)} \\ \sin(\beta)&=\frac35 \\ GB^2 &= AB^2+AG^2-2AB\times AG\cos(\beta) \\ AB &= \frac85 \end{aligned}$

Let l=AB. Draw a perpendicular GH from G to AB. Note that H is the midpoint of AB since GH is drawn perpendicular to AB. Moreover, triangles OGH & OFD are similar hence: $\frac{GH}{FD}$ = $\frac{3}{5}$ ; hence GH = $\frac{3}{5}$ since FD =1. Now,from right tr. AGH, $\frac{l^2}{4}$ =1 - $\frac{3^2}{5^2}$ or l = AB= $\frac{8}{5}$ .

Lets plot this figure on Cartesian Plane. $O$ will be Origin.

We get $G = (3,0)$ and $F = (5,0)$

Equation of circle with center $G$ will be $(x - 3)^2 + y^2 = 1$

Equation of Circle with center $F$ will be $(x - 5)^2 + y^2 = 1$

Let the tangent line meet the Circle ( $F$ centered) at $D(p,q)$

In traingle $ODF$ , using Phythagoras Theorem, we get $OD = \sqrt {5^2 - 1^2} = 2\sqrt6$

To calculate $p$ and $q$ , we have folllowing equations,

$p^2 + q^2 = 24$ and $(p - 5)^2 + q^2 = 1$

We get $q = \dfrac {\sqrt{24}}{5}$ and $p = \dfrac {24}{5}$

We get the equation of the tangent line as $x = \sqrt{24} y$

We now find the points of intersections of this line with $(x - 3)^2 + y^2 = 1$

We get two points as $B = (\sqrt{24}(\dfrac {3\sqrt{24} + 4}{25}) , \dfrac{3\sqrt{24} + 4}{25})$ and $A =(\sqrt{24}(\dfrac {3\sqrt{24} - 4}{25}) , \dfrac {3\sqrt{24} - 4}{25})$

We get $AB = \dfrac {8}{25} \sqrt{(\sqrt{24})^2 + 1^2} = \dfrac {8}{25}\times 5 = \dfrac {8}{5} = \boxed{\dfrac {8}{5} = \dfrac {a}{b} }$

Therefore, $a + b = 8 + 5 = \boxed{13}$

By comparing to similar triangle $\triangle ODF$ , we can see that the line from $G$ which bisects the chord $AB$ has a length of $\frac{3}{5}$ . Using Pythag, we can work out the length of $\frac{AB}{2}$ :

$\begin{aligned}(\frac{AB}{2})^2&=1^2 - 0.6^2\\ (\frac{AB}{2})^2&=1 - 0.36\\ (\frac{AB}{2})^2&=0.64\\ \frac{AB}{2} &= 0.8\\ AB =& 1.6 = \frac{8}{5}\\ 8+5 & =\boxed{13}\end{aligned}$

∆ OHG ~ ∆ ODF (AA similarity)

=> $\frac{OG}{OF}$ = $\frac{GH}{FD}$

We know OG = 3 units, OF = 5 units and FD = 1 unit => HG = $\frac{3}{5}$ units

If the perpendicular distance of a chord from the centre is given then length of the chord can be calculated easily using Pythagoras theorem.

=> AB= 2*sqrt(1- $\frac{9}{25}$ ) = $\frac{8}{5}$ .

From trigonometry on $\triangle DEF$ we have $OD = 2\sqrt{6}$ and $\cos O = \frac{2\sqrt{6}}{5}$ .

From the power of a point theorem , the two secants to circle $G$ have the relation $OA \cdot OB = 2 \cdot 4$ , or $OB = \frac{8}{OA}$ .

From the law of cosines on $\triangle OBG$ , $BG^2 = OB^2 + OG^2 - 2 \cdot OG \cdot OB \cdot \cos O$ , or $1^2 = (\frac{8}{OA})^2 + 3^2 - 2 \cdot 3 \cdot \frac{8}{OA} \cdot \frac{2\sqrt{6}}{5}$ , which solves to $OA = \frac{6\sqrt{6} \pm 4}{5}$ .

This makes $OB = \frac{8}{OA} = \frac{6\sqrt{6} \mp 4}{5}$ , which means $AB = OB - OA = \frac{8}{5}$ . Therefore, $a = 8$ and $b = 5$ , and $a + b = \boxed{13}$ .

Find below my solution

If you were told, "find a point $P$ on the line $OD$ satisfying $PG=1$ ," you would spot that there are two such points: $A$ and $B$ . This ambiguity is an example of the solution of a triangle $\Delta POG$ , knowing only two sides and a non-included angle (see here , for example). We have the following information about $\Delta POG$ :

• $OG=3$ (since the circles have unit radius)

• $PG=1$

• $\sin{\angle POG}=\frac{1}{5}$ (since $\angle POG=\angle DOF$ , $OF=5$ , $DF=1$ and $\angle ODF=90^{\circ}$ )

We're interested in the side length $OP$ ; let's call this $x$ . By the cosine rule, $PG^2 = OP^2+OG^2-2OP \cdot OG \cos{\angle POG}$ .

Substituting, this is $1^2=x^2+3^2-6x\cos{\angle POG}$ . Note that this is a quadratic in $x$ , with - as we'd expect - two solutions.

Rearranging and completing the square, this becomes $x^2-6x \cos{\angle POG} + 9\cos^2 {\angle POG} = -8+9\cos^2 {\angle POG}$

Plugging in $\sin{\angle POG}=\frac{1}{5}$ this becomes $(x-3 \cos{\angle POG})^2 = -8+9(1-\frac{1}{25}) = \frac{16}{25}$

Taking square roots, $x=3 \cos{\angle POG} \pm \frac{4}{5}$

The two solutions correspond to the lengths $OA$ and $OB$ ; we want $AB$ , the difference of these, which is just

$(3 \cos{\angle POG} + \frac{4}{5})-(3 \cos{\angle POG} - \frac{4}{5}) = \frac{8}{5}$ , so the required answer is $8+5=\boxed{13}$

The solution lies on our perspective as O is 1 and A is 2 B is 3 but actually I just a pythagorean theorem thingy 2^2 + 3^2 = 13 or c^2

Let X be the midpoint of AB. GX is perpendicular to AB.

GOX is similar to FOD, so GX=FD*GO/FO=3/5.

AX²=AG²-GX²=1²-(3/5)²=(4/5)².

AB=2AB=8/5.

Drop a perpendicular from G on chord AB, which can be easily proved=(3/5). Use Pythagoras to show

AB=2(√1-(9/25))=(8/5)

Answer=13

Let t be the mid-point of AB. Tan(<DOF) = 1/2.5 = .4. TG/OG = .4 = TG/1.5, TG = .6, TB = .8, AB = 1.6 = 8/5, 8 + 5 = 13. Ed Gray