Thread: Find the equation of the tangent to the circle

Find the equation of the tangent to the circle x^2 + y^ - 8y = 0 if the
point of tangency is (0,8).

My ATTEMPT: i solved for the center of the circle from the equation of the circle...
x^2 + y^2 - 8y = 0

x^2 + ( y - 4)^2 = 16
so, center ( 0, 4)
since point of tangency is ( 0,8) ...
im stuck here ...
because i wanted to find the slope of the segment (radius) joining the center and the point of tangency, so that after that I i would be able to solve the line perpendicular to the radius of the circle.

Can anybody help me on this please.

the tangent line to the circle will be perpendicular to the radius from the center to the point of tangency.

Originally Posted by skeeter

the tangent line to the circle will be perpendicular to the radius from the center to the point of tangency.

yeah.. so it is ok or possible to use the point-slope form for the equation?
m = 4/0 and P(0,8)
then
y - 8 = 4/0 (x - 0) ... hmmm what could be the equation of that line tangent to the circle.. can anybody show the solution please

because from the given answer of the book but has no solution is y = 8, may i know how it is done

make a sketch and you'll see why y = 8 is the equation of the tangent line ...

I think Skeeter's signature says it best - Work Smart, Not Hard. Surely if your centre is at (0, 4) and the point of tangency is (0, 8), then the radius needs to point UPWARDS, so is a vertical line. Therefore the tangent, being perpendicular to the radius, needs to be a HORIZONTAL line. Where is this horizontal line positioned?

GOt it BOYS! Love you Skeeter and Prove It

I need to get the points though that connect the radii and tangents to the circle, so that I can get the slope of each tangent and write the equation ...




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Originally Posted by rcs

yeah.. so it is ok or possible to use the point-slope form for the equation?
m = 4/0

This is your difficulty, isn't it? You cannot divide by 0, that does not give a number so a vertical line, like the line through (0, 4) and (0, 8) does NOT have a slope. However, it is easy to see that only horizontal lines are perpendicular to vertical lines and all horizontal lines have equation y= constant. Since we know the tangent line goes through (0, 8), that constant must be 8.

and P(0,8)
then
y - 8 = 4/0 (x - 0) ... hmmm what could be the equation of that line tangent to the circle.. can anybody show the solution please

because from the given answer of the book but has no solution is y = 8, may i know how it is done

you can see that the line goes through the point (3,4) and is perpendiular to the radius of the circle that also goes through the point (3,4).

the equation of the line tangent to the circle at the point (3,4) is:

y = -(3/4)*x + 25/4


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