Question

Find the area of the circle that has a center at point (2,3) and tangent to the line 4x - 3y = 5

Answer 1

Ok so lets put down what we know:
\[(x-2)^2+(y-3)^2=r^2\]
where r is the radius
this is the thing we want to find
if we can find r then we can find the area
\[A=\pi r^2\]
We also know that
the line 4x-3y=5 is tangent to the circle
So at some point, (a,b) , on the circle is the line tangent to the circle
Solving 4x-3y=5 for y we get
\[-3y=-4x+5\]
\[y=\frac{4}{3}x-\frac{5}{3}\]
Where 4/3 is the slope of this line
So when is y'=4/3
\[2(x-2)+2(y-3)y'=0\]
\[x-2+(y-3)y'=0 \text{ note: I divided both sides by 2}\]
\[(y-3)y'=-x+2\]
\[y'=\frac{-x+2}{y-3}\]
\[\frac{4}{3}=\frac{-a+2}{b-3} \text{ but gosh darn it we have two unknowns in this equation}\]
Remember at that point (a,b) did we have the slope is 4/3
Need more thinking
...

Answer 2

we also know that
\[(a-2)^2+(b-3)^2=r^2 \]
and \[b=\frac{4}{3}a-\frac{5}{3}\]

Answer 3

pi*(1.2)^2

Answer 4

lol wow derivatives in an analytic geometry question :P im *kind of* following though :D

Answer 5

derivatives?!!!

Answer 6

So
\[\frac{4}{3}=\frac{-a+2}{\frac{4}{3}a-\frac{5}{3}-3} \cdot \frac{3}{3}\]
\[\frac{4}{3}=\frac{3(-a+2)}{4a-5-9}\]
\[\frac{4}{3}=\frac{3(-a+2)}{4a-14}\]
\[4(4a-14)=3(-3a+6)\]
\[16a-56=-9a+18\]
\[25a=74\]
\[a=\frac{74}{25}\]
\[b=\frac{4}{3}a-\frac{5}{3}=\frac{4}{3}(\frac{74}{25})-\frac{5}{3}\]
\[b=\frac{296}{75}-\frac{125}{75}=\frac{171}{75}\]
So we have
\[(a-2)^2+(b-3)^2=r^2\]
\[(\frac{74}{25}-2)^2+(\frac{171}{75}-3)^2=r^2\]

Answer 7

\[\frac{36}{25}=r^2\]
\[r=\frac{6}{5}\]

Answer 8

So now you can find the area

Answer 9

yup

Answer 10

@A.Avinash_Goutham how did you get is so fast?
do you know a shorter way?

Answer 11

lol wow that sounds...uhmm long @_@

Answer 12

hmmm u just need the radius....it's the length of projection of that point on to that line

Answer 13

the radius is perpendicular to the tangent line :p

Answer 14

u got a formula for that ....
|ax+by+c|/(a^2 + b^2)^0.5

Answer 15

Calculus is sweet though :)

Answer 16

nd thanks for the medal :)

Answer 17

actually it's just 4x - 3y = 5

4x - 5 = 3y

slope is 4/3

that meas the slope of the radius is -3/4

since the center was given....

point slope form:

y - 3 = -3/4 (x - 2)
4y - 12 = -3x + 6
4y + 3x = 18

find the intersection

(3x + 4y = 18) 4
(4x - 3y = 5) -3

12x + 16y = 72
-12x + 9y = -15

im getting weird things now o.O

Answer 18

that's where i get confused :/

Answer 19

im trying to get the intersection of the radius and the tangent line so i can find the endpoint of the radius but it gets screwed up :C where did i go wrong @myininaya ?

Answer 20

you are doing good actually

Answer 21

almost there

Answer 22

Find (x,y)
then plug em into
\[(x-2)^2+(y-3)^2=r^2\]

Answer 23

yeah that's the problem cant find (x,y) :/ i get big numbers

Answer 24

y = mx + c nd m =-3/4 nd u have that point

Answer 25

\[25y=57\]
\[y=\frac{57}{25} \]

Answer 26

@lgbasallote did you get this for y?

Answer 27

yeah

Answer 28

\[3x+4(\frac{57}{25})=18\]

Answer 29

I plug this into one of the equations you gave

Answer 30

now we solve for x

Answer 31

\[3x+\frac{228}{25}=18\]
\[3x=18-\frac{228}{25}=\frac{222}{25}\]
\[x=\frac{1}{3} \cdot \frac{222}{25}=\frac{74}{25}\]

Answer 32

\[(\frac{74}{25}-2)^2+(\frac{57}{25}-3)^2=r^2\]

Answer 33

lol i just noticed the equation of the circle is also distance formula

Answer 34

\[\frac{576}{625}+\frac{324}{625}=r^2\]
\[\frac{900}{625}=r^2\]
\[\frac{36}{25}=r^2\]
lol @lgbasallote

Answer 35

6/5...oh wow they're the same *_*

Answer 36

i thought i got offcourse when i got big numbers haah

Answer 37

well there are more than one point on a circle lol

Answer 38

but i used the intersection of the tangent and the radius...

Answer 39

To add one more method:
4x - 3y = 5 => y = (4x-5)/3
(x - 2)^2 + (y-3)^2 = r^2 <--- put the value of y

(x - 2)^2 + ((4x-5)/3-3)^2 = r^2
Now you have quadratic equation in x ... if the line was tangent ... then there must be one and only one x ... and for that ... discriminant must be zero ...
i.e. b^2 - 4ac = 0 <---- there are no axes here .. only r, solve for r.

If you are doing analytical geometry .. you need this trick often!!

Answer 40

nahh im not doing analytic geometry :P you've seen my calculus problems...im just reviewing these for when i need them in calc...i suck at graphs =_=

Answer 41

seriously it sucks i can't give more than one medal :(

Answer 42

never mind ... just trade of tricks!!