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.

Re: Find the equation of the tangent to the circle

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

Re: Find the equation of the tangent to the circle

Quote:

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

Re: Find the equation of the tangent to the circle

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

Re: Find the equation of the tangent to the circle

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?

Re: Find the equation of the tangent to the circle

GOt it BOYS! Love you Skeeter and Prove It

Re: Find the equation of the tangent to the circle

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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Re: Find the equation of the tangent to the circle

Quote:

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.

Quote:

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

1 Attachment(s)

Re: Find the equation of the tangent to the circle

Attachment 26006

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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Re: Find the equation of the tangent to the circle

That last post is spam- it has nothing to do with the original question and was posted only to give that link.

Because the link is to a lawyer's website, from whom I would expect better things, I have notified them.