Question

Solve system of three equations. (from The Method of Coordinates, Russian Pre-Alg./Pre-Geo Book)

Answer 1

That's probably geometry, not pre alg

Answer 2

Solving three equations with three unknowns:
\[a^2+b^2=R^2\]
\[(c-a)^2 +b^2=R^2\]
\[(q-a)^2+(h-b)^2=R^2\]

Answer 3

I've found:
\[a = \frac{ c }{ 2 }\]

and

b = \[\frac{ q^2+h^2-cq }{ 2h }\]

Answer 4

The answer for R is given with

R = \[\frac{ \sqrt{(q^2+h^2)[(q-c)^2+h^2]} }{ 2h }\]

Answer 5

I'm not sure how R was gotten here.

Answer 6

Trying not to use trig. @robtobey

Answer 7

@freckles
@Hero

Answer 8

did you expand the last equation
and write the last equation in terms of q,h, and c?

Answer 9

I attempted to.

Answer 10

\[q^2-2a+a^2+h^2-2bh+b^2 = R^2\]

Answer 11

\[q^2-2qa+a^2+h^2-2bh+b^2=R^2 \\ (a^2+b^2)+q^2+h^2-2qa-2bh=R^2\]
you are missing a q on one of those terms

Answer 12

Okay thanks!

Answer 13

\[q^2+h^2-2qa-2bh=0\]

you said a can be written in terms of c
and also b can be written in terms of R and c
where R is what you want to solve for

Answer 14

Okay so I see that you can do that. I don't really know how that helps me to find R.

Answer 15

Ohhhh

Answer 16

So I have to write b in terms of R and c?

Answer 17

Does \[b = R + \frac{ c }{ 2 }\]

Answer 18

\[a^2+b^2=R^2 \\ (c-a)^2+b^2=R^2 \\ \implies \\ a^2+b^2=(c-a)^2+b^2 \\ a^2=(c-a)^2 \\ a=c-a \text{ or } a=-(c-a) \\ 2a=c \text{ or } c=0 \\ \text{ choosing the nontrivial case } a=\frac{c}{2} \\ \text{ now recall } a^2+b^2=R^2 \\ \text{ then } \frac{c^2}{4}+b^2=R^2 \\ b^2=R^2-\frac{c^2}{4} \\ b= \pm \sqrt{R^2-\frac{c^2}{4}} \\ \text{ assuming } b>0 \\ \text{ we have } b=\sqrt{R^2-\frac{c^2}{4}} \neq R+\frac{c}{2}\]

Answer 19

you should eventually wind up with
\[R=\sqrt{\frac{c^2 h^2+(q^2+h^2-qc)^2}{2h}} \text{ assuming } R>0\]
there is only a little more work to do to show this R is equivalent to the R you have above

Answer 20

oops type-o

Answer 21

\[R=\frac{\sqrt{c^2h^2+(q^2+h^2-qc)^2}}{2h}\]

Answer 22

\[(q^2+h^2-qc)^2=(q^2+h)^2-2(q^2+h^2)qc+q^2 c^2 \\ \text{ just using } (A+B)^2=A^2+2AB+B^2 \\ \text{ now we have } c^2h^2+(q^2+h^2-qc)^2 \\ =c^2h^2+(q^2+h^2)^2-2(q^2+h^2)qc+q^2c^2 \\ \\ =c^2(q^2+h^2)+(q^2+h^2)^2-2(q^2+h^2)qc \\ =(q^2+h^2)(c^2+q^2+h^2-2qc) \\ =(q^2+h^2)(c^2-2qc+q^2+h^2)\]
then one more thing to do to the inside of this square root thing

Answer 23

Thank you. I need to brush up on my algebra clearly.