question on properties of circle 9

ukorov · #1 $Old$ November 8th, 2009, 08:38 PM

Referring to the attached pic, find the radii of both circles.

Grandad · #2 $Old$ November 9th, 2009, 06:40 AM

Hello ukorov

Quote:

Originally Posted by ukorov $View Post$

Referring to the attached pic, find the radii of both circles.

In the attached diagram, I have marked the points of contact with the smaller circle as E and F.

Then, if the radius of the smaller circle is $x$

$O_1D=O_1F=DQ=FQ = x$

and if $PE = y$ , then $PD = y$ (tangents from a point to a circle are equal)

$PQ = x+y=6$

In $\triangle TPQ, TP= 10$ cm (Pythagoras)

$TE = TF \Rightarrow 10-y = 8-x \Rightarrow y-x=2$

Solving these simultaneous equations: $y = 4, x = 2$ .

So the radius of the smaller circle is $2$ cm.

Similarly let $PA = PC = w$ and $QC = QB = z$

Express the lengths of $TA$ and $TB$ in terms of $w$ and $z$ . Then use the fact that $PQ = 6$ to form another equation. Solve these equations to find $z$ , the radius of the larger circle.

Grandad

ukorov · #3 $Old$ November 9th, 2009, 10:48 AM

Quote:

Originally Posted by Grandad $View Post$

Hello ukorovIn the attached diagram, I have marked the points of contact with the smaller circle as E and F.

Then, if the radius of the smaller circle is $x$ [/size]

$O_1D=O_1F=DQ=FQ = x$

and if $PE = y$ , then $PD = y$ (tangents from a point to a circle are equal)

$PQ = x+y=6$

In $\triangle TPQ, TP= 10$ cm (Pythagoras)

$TE = TF \Rightarrow 10-y = 8-x \Rightarrow y-x=2$

Solving these simultaneous equations: $y = 4, x = 2$ .

So the radius of the smaller circle is $2$ cm.

Similarly let $PA = PC = w$ and $QC = QB = z$

Express the lengths of $TA$ and $TB$ in terms of $w$ and $z$ . Then use the fact that $PQ = 6$ to form another equation. Solve these equations to find $z$ , the radius of the larger circle.

Grandad

PQ = 6 = w + z
w = 6 - z .....(1)
TA = TP + w = 10 + w
TB = TQ + z = 8 + z
hence 10 + w = 8 + z .....(2)
(1) to (2):
10 + 6 - z - 8 = z
2z = 8
z = 4
hence larger circle has radius 4cm.

Thanks