Determine equation of circle

Question

34. (a) Find the equation of AB, the
perpendicular bisector of the line
joing the points P(−3, 1) and
Q(1, 9).

(b) C is the centre of a circle passing
through P and Q. Given that QC is
parallel to the y-axis, determine the
equation of the circle.

(c) The tangents at P and Q intersect at

Write down
(i) the equation of the tangent at Q
(ii) the coordinates of T.

Answer 1

a)  perpendicular bisector will go through the mid point of the given line segment,

     to be perpendicular it will have SLOPE = negative reciprocal slope of PQ

 

line segment is from P(-3,1) to Q(1,9).   it has a mid-pt of  (X1 + X2)/2,  (Y1 + Y2)/2 

or  X = (-3 + 1)/2 = -1    and Y = (1 + 9)/2  = 5

==> Midpt is then (-1,5)

 

slope of the line segment =  (Y2 - Y1)/ (X2 - X1) 

slope =  (9-1)/(1-(-3)) = 8/4 = 2

==> it's slope is the neg reciprocal = -1/2

 

for slope, intercept form of a line:  Y = mX + B

using m = -1/2 and plugging in the mid-pt found above (-1,5), we then solve for B

     So using Y = mX + B, we get   5 = -1/2*(-1) + B

 so B = 5 - 1/2 = 4.5, 

SO, THE PERP BISECTOR LINE EQN  =  Y = -1/2X + 4.5  OR  Y = -0.5X + 4.5

 

 

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b)  equation of a circle, with center H,K and radius R 

     (X-H)^2  + (Y-K)^2  = R^2

 

     if the QC is parallel to the Y axis that means it has the same X value as Q

     so we know the center H,K is  1,K   (H=1, K is unknown)

     

     substituting our two pts, one at a time in the circle eqn

     

    1)   (-3 - 1)^2  +    (1 - K)^2          = R^2      plugging in  P = -3,1  and H=1

          or   16       +  (1 - 2K + K^2)    = R^2

 

         rearranging    ===>  K^2 - 2K + 17 = R^2

 

    2)   (1 - 1)^2  +  (9 - K)^2  = R^2      plugging in  Q =  1,9  and H=1

               0         +  81 - 18K + K^2 = R^2

 

         rearranging    ===>  K^2 - 18K + 81  = R^2

 

    BOTH EQUATIONS = R^2,   SO, SET THEM EQUAL TO EACH OTHER

 

          K^2 - 2K + 17  =  K^2 - 18K + 81        (K^2 terms cancel out)

                

          -2K+18K = 64  or   16K  = 64    OR  ===> K = 4

 

   SO, center of circle is H,K = 1,4

 

   plugging this value of H,K and either point (since they are both on the circle) in the circle equation, 

   we can solve for the radius R.

 

   (X-H)^2 + (Y-K)^2 = R^2

 

   if we use point Q = (1,9) and we found the center pt  H,K = (1,4)

 

   (1 - 1)^2 + (9 - 4)^2 = R^2

 

        0        +    25  = R^2  

 

    ====>  R = 5     (=radius)

 

    THE CIRCLE EQN IS THEN =   (X-1)^2 + (Y-4)^2 = 25  

 

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C)  Tangents at P,Q intersect at  ?

 

     Tangents to the circle at P,Q will be lines perpendicular to the line from the point to the center point.

     to be perpendicular it is the negative reciprocal slope 

 

     the radial line PC has a slope, m =  (Yp - K)/(Xp - H)       [ P =(Xp,Yp)  C=(H,K) ]

                                                         (1 -  4)/(-3 - 1)           P =(-3,1)    C= (1,4)

     m = -3/-4 = 3/4  (slope of line PC)

 

     the TANGENT LINE is then the neg reciprocal slope of the PC, or -4/3

     plugging in P = (-3,1) in the pt slope eqn of the line

     Y = mx + B

 

     1 = -4/3*(-3) + B   so B = 1-4 = -3

 

     SO THE TANGENT LINE TO P is then:

     ===>     Y = -4/3X - 3

 

     THE OTHER TANGENT LINE AT Q, is found the same way

     the radial line QC has a slope, m = (Yq - K)/(Xq - H)        [ Q =(Xq,Yq) C=(H,K) ]

                                                 m = (9 - 4)/(1 - 1)               Q=(1,9) C= (1,4)
     BUT !  Slope is undefined because we divide by zero, and we already knew it was 

     parallel to the Y axis with a X value = 1  (remember, parallel to Y axis and goes through Q=1,9)

     so the line eqn is just ==>    X=1

 

     THE INTERSECTION POINT OF THE TWO TANGENT LINES 

     Y = -4/3X - 3   and X = 1, we just plug in X = 1 in the first line eqn, OR

 

     Y = -4/3*1 - 3  so the Y intersection point is = -13/3 or -4.33

 

     INTERSECTION POINT IS (1,-4.33)