<ol>
<li>Which of the following is the equation of the circle that has its center at the origin and is tangent to the line with equation 3x-4y=10?
A)x^2+y^2=2
B)x^2+y^2=4
C)x^2+y^2=3
D)x^2+y^2=5
E)x^2+y^2=10</li>
</ol>
<p>Barron's answer key tells me to use a formula...but it doesn't tell me what the formula is.</p>
<ol>
<li>The graph of xy - 4x - y - 4 = 0 can be expressed as a set of parametric equations. If y = 4t/(t-3), and x=f(t), then f(t) =
A)t + 1
B)t - 1
C)3t - 3
D)(t - 3)/4t
E)(t - 3)/2</li>
</ol>
<p>I know I have to substitute 4t/(t -3) for y in the original equation...but it gets messy and I can't seem to get the answer. Could someone please show me all the steps?</p>
<ol>
<li><p>Draw a picture. The problem reduces down to finding the distance from (0,0) to the line 3x-4y-10 = 0. If you know the distance formula, the distance is |10|/sqrt(3^2 + (-4)^2) = 2, so the radius of the circle is 2. Since its center is at the origin, the equation is x^2 + y^2 = 4.</p></li>
<li><p>For some reason I got x = (2t - 3)/3, which doesn’t march the above choices. I factored the equation to (x-1)(y-4) = 8, and y-4 = 12(t-3). Then x-1 = 2(t-3)/3, x = (2t - 3)/3.</p></li>
</ol>
<p>Edit: Try plugging t=0, then y=0. Solving yields x = -1 (at t=0). Only choice B satisfies this. If we let t=4 so that y=16, we see that x = 5/3 at t = 4. No answer choice satisfies this, but x = (2t - 3)/3 does.</p>