SOLUTION: a. Show that the point Q(6,2) lies on the circle x^2+y^2-4x+2y-20 b. Find the equation of tangent to the circle at point Q

Algebra ->  Circles -> SOLUTION: a. Show that the point Q(6,2) lies on the circle x^2+y^2-4x+2y-20 b. Find the equation of tangent to the circle at point Q      Log On


   

Question 1129332: a. Show that the point Q(6,2) lies on the circle x^2+y^2-4x+2y-20
b. Find the equation of tangent to the circle at point Q
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
a. Show that the point Q(6,2) lies on the circle
x%5E2%2By%5E2-4x%2B2y-20
6%5E2%2B2%5E2-4%2A6%2B2%2A2-20
36%2B4-24%2B4-20
44-44
0=>the point Q(6,2) lies on the circle

b. Find the equation of tangent to the circle at point Q

As a tangent is a straight line it is described by an equation in the form
+y+-+y%5B1%5D+=+m%28x+-x%5B1%5D%29.
You need both a point and the slope to find its equation.
You are usually given the point - it's where the tangent meets the circle.
To find the slope use the fact that the tangent is perpendicular to the radius from the point it meets the circle.
Work out the slope of the radius (r) at the point the tangent meets the circle. Then use the equation +m_tg+=+-+1%2Fm_r to find the gradient of the tangent.
%28x%5E2-4x%29%2B%28y%5E2%2B2y%29=20...if you complete squares, you will see that the center of the circle is (2,-1)
%28x%5E2-4x%2Bb%5E2%29-b%5E2%2B%28y%5E2%2B2y%2Bb%5E2%29-b%5E2=20
%28x%5E2-4x%2B2%5E2%29-2%5E2%2B%28y%5E2%2B2y%2B1%5E2%29-b%5E2=20
%28x-2%29%5E2%29-4%2B%28y%2B1%29%5E2%29-1=20
%28x-2%29%5E2%29%2B%28y%2B1%29%5E2%29=25
Q(6,2) and (2,-1)=>
the slope of the radius
m_r=%28-1-2%29%2F%282-6%29=-3%2F-4=3%2F4
the slope of the tangent is m_tg=-1%2F%283%2F4%29=-4%2F3
and equation af the tangent is:

Q(6,2) ,m_tg=-4%2F3
y+-+2+=+-%284%2F3%29%28x+-6%29
y++=+-%284%2F3%29x+-6%28-4%2F3%29%2B2
y++=+-%284%2F3%29x+%2B8%2B2
y++=+-%284%2F3%29x+%2B10