We are given that the common chord AB between two circles, centered at P and centered at Q, is perpendicular to the segment connecting the centers of the circles. Let's take a look at the given diagram.
First notice that P and Q are equidistant from the endpoints of AB. Next, recall that if a radius of a circle is perpendicular to a chord, then it bisects the chord and its arc. Therefore PQ is a perpendicular bisector of AB.
If we name the point of intersection of AB and PQ with the variable S, then we can say that AS=SB. Since we are given that AB=10 each of the segments of AB has a length of 5.
Now we can find PS and SQ using the Pythagorean Theorem. According to this theorem the sum of the squared legs of a right triangle is equal to the squared hypotenuse. Let's write the equations for PS and SQ.
PS^2+5^2=11^2
SQ^2+5^2= 9^2
Let's solve the above equations. Notice that since PS and SQ represent the lengths, we will consider only the positive cases when taking a square root of these values.
lcPS^2+5^2=11^2 & (I) SQ^2+5^2=9^2 & (II)
(I), (II): Calculate power
lPS^2+25=121 SQ^2+25=81
(I), (II): LHS-25=RHS-25
lPS^2=96 SQ^2=56
(I), (II): sqrt(LHS)=sqrt(RHS)
lsqrt(PS^2)=sqrt(96) sqrt(SQ^2)=sqrt(56)
(I), (II): sqrt(a^2)=a
lPS=sqrt(96) SQ=sqrt(56)
(I), (II): Use a calculator
lPS=9.7979... SQ=7.4833...
(I), (II): Round to 2 decimal place(s)
lPS≈ 9.80 SQ≈ 7.48
We found that PS and SQ are approximately 9.80 and 7.48. Finally, according to the Segment Addition Postulate we will add add these segment lengths to find the length of PQ.
PQ=PS+SQ=9.80+ 7.48≈ 17.3
