The equation of a circle is (x + 6)^2 + (y - 4)^2 = 16. the point (-6, 8) is on the circle.

what is the equation of the line that is tangent to the circle at (-6, 8)?

y = 8 is the equation of tangent.

Step-by-step explanation:

The equation of the tangent to the circle at (-6,8) is of the form:

y = mx + c

where m is the slope of the tangent and c is the y-intercept.

The point (-6,8) lies on the circle and the tangent line as well.

Hence (-6,8) satisfies the line equation:

8 = m(-6) + c ⇒ c-6m = 8 -------------1

We know that slope of two perpendicular lines are related as:

At any point on the circle, the normal line at a point is always perpendicular to the tangent line at that point.

Hence :

We can find the slope of the normal at point (-6,8) as it passes through the centre of the circle (-6,4) by using the two-points formula for slope.

          = ∞

Slope of the normal is infinity and hence slope of tangent is -1/∞ = 0

Hence m=0

Putting m=0 in equation 1 we get:

c = 8

The equation of tangent line at (-6,8) is:

y = 8

i think it will be zero as slope has a formula of difference between initial and final distance divided by itz time interval