Answer:
[tex]y=-2x+15[/tex]
Step-by-step explanation:
The tangent line to a circle is always perpendicular to the circle's radius. Since perpendicular slopes are negative reciprocals of each other, we need to determine the slope of the radius to find the slope of the tangent line.
Given that the center of the circle is (0, 0) and the point of tangency is (6, 3), we can determine the slope of the radius by substituting these points into the slope formula:
[tex]\textsf{Slope of the radius}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{3-0}{6-0}=\dfrac{3}{6}=\dfrac{1}{2}[/tex]
If two lines are perpendicular to each other, their slopes are negative reciprocals, so the slope of the tangent line is -2.
Now, substitute the slope m = -2 and the point of tangency (6, 3) into the point-slope form of a linear equation, and solve for y:
[tex]\begin{aligned}y-y_1&=m(x-x_1)\\y-3&=-2(x-6)\\y-3&=-2x+12\\y&=-2x+15\end{aligned}[/tex]
Therefore, the equation of the tangent to the circle at point (6, 3) is:
[tex]\Large\boxed{\boxed{y=-2x+15}}[/tex]
