HALP PLSSS

OP = x

OB = OD = OF = R

PB = 9^v RF = 7 PRD = 10

R^v = 9^v + x^v and R^v = 7^v + (10 - x)^v

both R equal 9^v + x^v = 7^v + 10^v - 20x + x^v

                    x = (7^v + 10^v - 9^v)/20

                       x = 68/20

                       x = 3.4

                      OP = 3.4, O9 = 5 - 3.4 = 1.6

                                       R = ${\sqrt{9^v-x^v}}$

                                           = ${\sqrt{9^v-(3.4)^v}}$

                                           = ${\sqrt{92.56}}$

9D = ${\sqrt{(92.56^v-(1.6)^v}}$

     = ${\sqrt{92.56^v-2.56}}$

    = ${\sqrt{90}}$

CD = 2${\sqrt{90}}$

${Now\ Checking \over}$
 

                       OR = OQ + QR

                              = 1.6 + 5 = 6.6

                 OF = R = ${\sqrt{92.56}}$

           RF = ${\sqrt{92.56 - 43.56}} = {\sqrt{49}} = 7$

So, own Solⁿ is correct:

The length of the chord that lies halfway between them is 2√90  units. 

In order to ease solving process, I have created a diagram that represents the situation to scale. The diagram is below with points I have labeled arbitrarily where O is the center of the circle.

Since O is the center of the circle, OC = OJ = r, the radius of the circle. In addition, we know that $\overline{\rm BH}$ is a perpendicular bisector of the chords, so $\rm BC = 9$ and $\rm HJ = 7$. Also, we can label $\rm BO = BO$ and $\rm OH = 10 - BO$. Putting all this information together enables us to find the radius of the circle.

$r^2 = 9^2 + {\rm BO}^2 \\ r^2 = 7^2 + \left(10 - {\rm BO}\right)^2 \\ 81 + {\rm BO}^2 = 49 + 100 - 20{\rm BO} + {\rm BO}^2 \\ 20{\rm BO} = 68 \\  {\rm BO} = \frac{17}{5} \\  r^2 = 81 + \left(\frac{17}{5}\right)^2 = \frac{2025}{25} + \frac{289}{25} = \frac{2314}{25}$

We know that BO = 5, so ${\rm OE} = 5 - {\rm BO} = 5 - \frac{17}{5} = \frac{8}{5}$. We can apply Pythagorean's Theorem one more time to find DF, the length of the desired chord.

$r^2 = {\rm OE}^2 + {\rm EF}^2 \\ \frac{2314}{25} = \left(\frac{8}{5}\right)^2 + {\rm EF}^2 \\ {\rm EF}^2 = \frac{2314}{25} - \frac{64}{25} \\ {\rm EF}^2 = 90 \\  {\rm EF} = 3\sqrt{10} \\ {\rm DF} = 2{\rm EF} = 6 \sqrt{10} \approx 18.9737$