# 2013 MTAP Saturday Class Challenge Problem Solution Part 2

Previously, we solved problems 1 to 3 of the MTAP Saturday Class S-5 for 4th year. Now let’s continue the problem solving with the rest of the problems.

Problem 6:

If the length of the common chord intersecting circles is $16$ feet and the radii are $10$ and $17$ feet, find the distance between the centers of the circle.

Solution: Draw  and label the figure as shown below

Since $FG=16, FH=HG=8$

Observe that $\triangle FBH$ and $\triangle FHD$ are right triangles, thus we can use Pythagorean Theorem to solve for $BH$ and $DH$.

Solving for $BH$:

$10^2=8^2+BH^2$

$100=64+BH^2$

$BH^2=100-64=36$

$BH=6$

Solving for $DH$:

$17^2=8^2+DH^2$

$289=64+DH^2$

$DH^2=289-64=225$

$DH=15$

The distance between the circles $BD=BH+DH$

Solving for the distance between two circles we have,

$BD=BH+DH$

$BD=6+15$

$\boxed{BD=21}$

Problem 7:

The shortest distance between from a point to a circle is 6 in. and the greatest distance from the same point to a circle is 24 in. Find the length of the tangent from that point.

Solution: Draw and label the figure accordingly,

In the figure, PA is the shortest distance and PB is the greatest distance. PT is the length of the tangent. We assign it by a variable x.

$AO=OT=OB=9$ since they are all the radii of the circle.

$\triangle PTO$ is a right triangle with right angle at $\angle{PTO}$.

Using Pythagorean Theorem in $\triangle PTO$ we have,

$PO^2=PT^2+OT^2$

$15^2=x^2+9^2$

$x^2=15^2-9^2=144$

$\boxed{x=12 in}$

Problem 8:

From an external point two secants, PAB and PCD, are drawn, one cutting the circle at A and B, the other cutting it at C and D. If PB=27 in., AB= 21 in., and PC=CD. What is the length of PD?

Solution: Draw and label the figure accordingly as shown below

Using the secant-secant theorem,

$PA\cdot PB=PC\cdot PD$

$6\cdot 27=x\cdot 2x$

$2\cdot 9^2=2x^2$

$x^2=9^2$

$x=9$

Since $PD=2PC,PD=2\cdot9=\boxed{18 in}$