the length of PQ

Here's another approach....find the circle's center as Alan did  = (5,11)

And the distance from a line to a point not on that line is given by

d = abs ( Ax + By + C) / √(A² + B²)       where (x,y) is the given point and Ax + By + C = 0 is the equation of the line.....so we have....

d = abs (12(5) + 5(11) + 54 ) / √(12² + 5²) = abs(169) / 13 = 13

Here's the graph ...

GRAPH

1.  Write the circle equation in the form (x-xc)² + (y-yc)² = r²,  where (xc, yc) are the coordinates of the centre and r is the radius.  

2. Your circle equation can be written as (x-5)² + (y-11)² = 10², so the centre is at (5, 11).

3. Your straight line can be written in the form y1 = -(12/5)x - 54/5, so it has slope -12/5.

4. The slope of the straight line perpendicular to this is y2 = (5/12)x + k, where k is a constant.

5. For this second line to go through (5, 11) we must have 11 = (5/12)*5 + k so that k = 107/12

6. The equation of the second line is therefore y2 = (5/12)x + 107/12

7. This hits the first line when y1=y2.  The value of x at which this occurs is given by equating the two straight line equations:  -(12/5)x - 54/5 = (5/12)x + 107/12

8.  Rearrange as;  (5/12 + 12/5)x = -54/5 - 107/12  or (169/60)x = -1183/60 so that x = -1183/169 = -7

9. When x = -7 then y1 = y2 = -(5/12)*7 +107/12 = 72/12 = 6

10. The coordinates of P are therefore (-7, 6)

11. The distance between P and Q is given by √( (5 - (-7))² + (11 - 6)² ) = 25√2 ≈ 35.355

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Oops!  Ignore my result for 11.  It should be √( (5 - (-7))² + (11 - 6)² ) =√( (12²+5²) = √169 = 13. (thanks Chris!).

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Very nice, Alan....!!!

I really like that one....a definite "Daily Wrap" candidate