Question

Given the line 3x-4y-12=0, find the equation of the circle with center at (1,1) and tangent to the line.

Answer 1

do you know the distance formula? , @iamforeveryoung

Answer 2

Yeah...d=square root of (xsub2-xsub1)^2+(ysub2-ysub1)^2

Answer 3

like this way : d( centre, L)=|3X1-4X1-12| / sqrt(3^2+4^2)

Answer 4

this is the distance between a point and a line

Answer 5

Oh.ok.thanks for the help :)

Answer 6

do you know what to do next?
sorry I am not sure if I did help
@iamforeveryoung

Answer 7

Nahh..i was asking some of my classmates but they also don't know,what should i do next?

Answer 8

The distance between a point \((x_0,y_0)\) and a line in standard form \(Ax+By=C\) is given by \[d=\frac{|Ax_o+By_o+C|}{\sqrt{A^2+B^2}}\]

The distance to the tangent line will be the radius of the circle. Do you know the formula for a circle at an arbitrary center?

Answer 9

sorry for keeping you waiting,
the distance will be the radius of the circle
and the formula of the circle will be
(x-1)^2+ (y-1)^2=r^2

Answer 10

@whpalmer4 thanks for posting the formula,
but i think it should be this way Ax+By+C=0, is it?

Answer 11

Sorry, I think that formula would be -C, not +C...

Answer 12

(my formula, that is)

Answer 13

@iamfoerveryoung? is it ok ?

Answer 14

@ranxu6j3 sorry, I couldn't decipher your posting of the formula, that's the only reason I posted...

Answer 15

iamfoeveryoung, because the line is tangent to the circle, the distance between the center and the line would be the radius

Answer 16

Here's a hint that won't be much help :-) point of contact is at (64/25,-27/25)

Answer 17

@whpalmer4 , that's very ok , thanks for doing that ,cuz i am not good at using the tool of equation,
and i've checked my books and internet , about the formula: should be+c
check this : http://csm01.csu.edu.tw/0166/Math1/32/321.htm

Answer 18

Yes, I was thinking of the other standard form of a line, which is Ax + By = C.

Standards are wonderful, everyone should have one :-)

Answer 19

@ranxu6j3 and @whpalmer thanks for the help, i'll try to understand everything you posted :)

Answer 20

@whpalmer thanks:)