Question

Draw the unit circle with center (0,0)and plot the point P=(6,2). Observe there are TWO lines tangent to the circle passing through the point P.
1)the tangent line on top of the circle(L1) has coordinates?

a) what is the equation for L1?

2) the tangent line below the circle(L2) has coordinates?

b)what is the equation for L2?

Answer 1

Answer:

Step-by-step explanation:

Given that a tangent is drawn from P (6,2) to unit circle with center at the origin.

The tangent passing through (6,2) would have equaiton of the form

$y-6 = m(x-2)\\y =mx-2m+6$ for suitable m.

Because this line is tangent, distance of centre of circle namely origin is the radius 1.

i.e. $|\frac{-2m+6}{\sqrt{1+m^2} } |=1\\(-2m+6 )^2 = 1+m^2\\3m^2-24m+35=0\\\\m=1.918, 6.082$

the coordinates are the intersection of the tangent with the circle

They are (-0.162, 0.987) and (0.462,-0.887)

Answer 2

Firstly, We created a pencil outline initialized color and two lines to scale. Now, again for the line that passes through, solve the following.

$\to P(6,2) \\\\\to y-2=m(x-6)\\\\ \to m=slope$

• Solve the following for the perpendicular line to the first arc and go through (0,0).

$\to y=-(\frac{1}{m})x$

• We have a solution if you solve for m.

$\to m=-\frac{y}{x}$

• In the first equation, we have and sub

$\to y-2=-(\frac{y}{x})(x-6)$

• Now that y and x are on the unit circle,

$\to x^2+y^2=1$

• When you solve these two equations for x and y, you will find (messy)

$x=-.16225\\\\ y=.98675 \\\\x=.46225\\\\ y=-.88675$

• The very first set is the dimensions of the point of tangency on top, and also the second set seems to be the coordinates of the equivalent lower point.

Given that you have two points on each line, you can easily write the equations for the lines or compute the gradient m directly. That it was a fun, out-of-the-ordinary issue.

Learn more:

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