SOLUTION: The point P moves along an arc of a circle with centre E(2, 3). the arc of the circle passes through A(-2, 0) and B(5, k). (a) Find (i) the equation of the locus of the point P,

Algebra ->  Formulas -> SOLUTION: The point P moves along an arc of a circle with centre E(2, 3). the arc of the circle passes through A(-2, 0) and B(5, k). (a) Find (i) the equation of the locus of the point P,      Log On


   

Question 934410: The point P moves along an arc of a circle with centre E(2, 3). the arc of the circle passes through A(-2, 0) and B(5, k).
(a) Find
(i) the equation of the locus of the point P,
(ii) the values of k.
(b) The tangent of the circle at the point A intersects the y-axis at the point Q. Find the area of the triangle OAQ where Of is the origin.

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The general equation of a circle is,
%28x-h%29%5E2%2B%28y-k%29%5E2=R%5E2
%28x-2%29%5E2%2B%28y-3%29%5E2=R%5E2
Use A to get R.
%28-2-2%29%5E2%2B%280-3%29%5E2=R%5E2
16%2B9=R%5E2
R%5E2=25
R=5
.
.
.
%28x-2%29%5E2%2B%28y-3%29%5E2=25
%285-2%29%5E2%2B%28k-3%29%5E2=25
9%2B%28k-3%29%5E2=25
%28k-3%29%5E2=16
k-3=+0+%2B-+4
k=3+%2B-+4
k=-1 and k=7
.
.
.
2%28x-2%29dx%2B2%28y-3%29dy=0
%28y-3%29dy=-%28x-2%29dx
dy%2Fdx=-%28x-2%29%2F%28y-3%29
At A,
dy%2Fdx=-%28-2-2%29%2F%280-3%29
dy%2Fdx=-%284%2F3%29

The value of the derivative is equal to the slope of the tangent line.
Using the point-slope form of a line,
y-0=-%284%2F3%29%28x-%28-2%29%29
y=-%284%2F3%29%28x%2B2%29
So when x=0
y=-%284%2F3%29%282%29
y=-8%2F3
So know you know the base and height of the triangle.
A=%281%2F2%29%282%29%288%2F3%29
A=8%2F3