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
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-> 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
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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) (Show Source):
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 .
You need both a and the to find its equation.
You are usually given the point - it's where the tangent meets the circle.
To find the use the fact that the tangent is to the from the point it meets the circle.
Work out the slope of the (r) at the point the tangent meets the circle. Then use the equation to find the gradient of the tangent.
...if you complete squares, you will see that the center of the circle is (,)
Q(,) and (,)=>
the slope of the radius
the slope of the tangent is
and equation af the tangent is: