Given two circles with radii $8$ and $6$ units with centers $A$ and $B$ such that $AB=12$ If $P$ is mid point of $QR$
Find Length of $QP$
My try:
I assumed $A(0,0)$ and $B(12,0)$
So the equations of circles are:
$$x^2+y^2=64$$
$$(x-12)^2+y^2=36$$
Solving above equations we get:
$$P\left(\frac{43}{6}, \frac{\sqrt{455}}{6}\right)$$
Then we can assume
$$Q(8\cos\alpha, 8\sin \alpha)$$
$$R(12+6\cos\beta,6\sin\beta)$$
Using mid point formula we get:
$$8\cos\alpha+6\cos\beta=\frac{7}{3}$$
$$8\sin\alpha+6\sin\beta=\frac{\sqrt{455}}{3}$$
Any idea here?
You don't need any trigonometry or analytical geometry. The following equations are quite obvious (Pythagora):
$$p^2+\frac{x^2}{4}=8^2=64\tag{1}$$
$$q^2+\frac{x^2}{4}=6^2=36\tag{2}$$
$$(p-q)^2+x^2=12^2=144\tag{3}$$
Introduce substitution: $y=\frac{x^2}{4}$ and you get:
$$p=\sqrt{64-y}\tag{4}$$
$$q=\sqrt{36-y}\tag{5}$$
$$(p-q)^2+4y=144\tag{6}$$
Replace (4) and (5) into (6):
$$(\sqrt{64-y}-\sqrt{36-y})^2+4y=144$$
$$100-2y-2\sqrt{(64-y)(36-y)}+4y=144$$
$$2y-44 = 2\sqrt{(64-y)(36-y)}$$
$$y-22=\sqrt{(64-y)(36-y)}$$
$$(y-22)^2=(64-y)(36-y)$$
$$(y-22)^2-(64-y)(36-y)=0$$
$$56y-1820=0$$
$$y=\frac{x^2}{4}=\frac{1820}{56}=\frac{65}{2}$$
$$x=\sqrt{130}$$