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

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

anonymous · Answer

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...