Question

What is the equation of the circle which is tangent to the line 3x-4y=32, and the center is at point (0,7)

Answer 1

@ganeshie8 pls help sir

Answer 2

equation of circle whose center is \((h,k)\) and radius is \(r\) :

\[(x-h)^2 + (y-k)^2 = r^2\]

Answer 3

@ganeshie8 what will happen next?

Answer 4

familiar with `point to line` distance formula ?

Answer 5

http://wpcontent.answcdn.com/math/5/a/e/5aea3e408bf7348f5e1720d4b8043fcf.png

Answer 6

you may use that formula to find the `radius`, which is the distance between `center` and the `tangent line`

Answer 7

line : 3x-4y-32 = 0
point : (0, 7)

radius = \(\large \dfrac{|3(0) - 4(7) - 32|}{\sqrt{3^2+4^2}} = ?\)

Answer 8

\[\frac{ 60 }{ \sqrt{25} }\]

Answer 9

\[\large \dfrac{|3(0) - 4(7) - 32|}{\sqrt{3^2+4^2}} = \dfrac{60}{\sqrt{25}} = \dfrac{60}{5} = 12\]

Answer 10

sorry divided the wrong number. xD

Answer 11

:) plugin center and radius in the equation of circle @Faye281 still here ?

Answer 12

@ganeshie8 would (x-0)^2+(y-7)^2=144

Answer 13

x^2+y^2-14y+49=144?

Answer 14

@ganeshie8 sorry. Afk

Answer 15

that looks good ^

Answer 16

@ganeshie8 sorry :( x^2+y^2-14y+49=144? is the answer? aren't we going to transfer 144 to the left equation?