Question

Write the equation of the line tangent to the circle at the given point.

x2 + y2 = 13 at (2, 3)

Question 14 options:

y=−23x−133

y=−23x+133

y=23x−133

y=23x+133

Answer 1

The circle's origin at (0,0) and radius \(r=\sqrt{13}\)... a line from the origin to (2,3) is perpendicular to the line tangent to the circle at given point... and the slope from (0,0) to (2,3) is \[m=\frac{0-3}{0-2}=\frac{3}{2}\]so the line tangent to the given circle will have a slope \[m_1=- \frac{1}{m_2}\] for perpendicular lines... so if \(m_2=m=3/2\), then \(m_1=-2/3\).
Since a point (2,3) is the tangent point of the tangent line, by point-slope form
\[(y-y_1)=m(x-x_1)\]\[(y-3)=-\frac{2}{3}(x-2)\]\[3(y-3)=-2x+4\]\[3y-9=-2x+4\]\[3y=-2x+4+9=-2x+13\]\[y=-\frac{2}{3}x+\frac{13}{3}\]
it seems the second one...