Start by using the slope formula to find the slope of the line passing through the center of the circle and the point (5,6). Then, take the opposite reciprocal of this to find the slope of the line tangent to the circle.
x-intercept: (13,0) y-intercept: (0,39/4)
Practice makes perfect
We first need to find the equation of the tangent line that is perpendicular to the radius at the point (5,6). To find the equation of the line we first need to identify the center of the circle. The standard form for the equation of a circle with center ( h, k) is (x- h)^2+(y- k)^2= r^2.
Let's look at the given equation.
(x- 2)^2+(y- 2)^2= 5^2
We see above that h= 2, k= 2, and r= 5. Therefore, the center of the circle is at ( 2, 2) and has a radius of 5. Let's graph this circle and sketch a line tangent to the circle at the point (5,6).
Let's use the formula for slope to find the slope of the line passing through the two points — the center ( 2, 2) and the point (5,6).
m = y_2- y_1/x_2- x_1=6- 2/5- 2= 4/3
The line connecting the two points will be perpendicular to the tangent line. Since perpendicular lines have opposite reciprocal slopes, the slope of the tangent line is - 34. Using the point-slope form of a line, we can find the equation of the tangent line.
y-y_1=m(x-x_1)
Let's substitute (x_1,y_1)=(5,6) and m=- 34 to solve for the equation of the tangent line in the form y=mx+b.
The x-intercept of the tangent line is (13,0). To find the y-intercept we could repeat this process and substitute x=0 to solve for y. However, because our equation of the tangent line is already in slope-intercept form, the y-intercept will be equivalent to the value for b.
y=mx+b
⇕
y=-3/4x+39/4.
The y-intercept of the tangent line is (0, 394).
