Question

determine the equation of the circle which passes through the point (5,4) and is tangent to the y-axis at the point (0,2)

ok so i know how an equation of a circle should look

Ax^2+Ay^2+Bx+Cy+D=0

and tangent to the y-axis...is what is triggering me off because I would just use the distance formula if it was worded otherwise so....what do i do?

Answer 1

Tangent to Y axis suggest that the point must be at the half height

Answer 2

huh?

Answer 3

This show tangent line at various point

Answer 4

oh okay

Answer 5

so its all equal to 2^2...but...

Answer 6

We can plug in two point
(5,4) which is given
(0,2) which is tangent,
(0-a)^2+(2-b)^2
(5-a)^2+(4-b)^2

Answer 7

now i am confused. if we agree that the center must be at (b,2) than doesn't it follow that
\[b=(b-5)^2+(4-2)^2\]? that is the distance between (b,2) and (5, 4) is the same as the distance between (0,2) and (b,2)?

Answer 8

try as i might i cannot draw a circle tangent to the y-axis at (0,2) that does not have center at (b,2) for some b

Answer 9

and if i do i get the center at
\[\frac{11+\sqrt{5}}{2}\] which is a perfectly good number but seems odd in this context

Answer 10

i know we have to use the distance formula but..."is tangent to the y-axis at the point (0,2)"

Answer 11

myininaya found the derivative in terms of y'

Answer 12

when i draw a circle tangent to the y- axis at (0,2) is surely looks like the center must be at (b,0) i maybe mistaken, but i cannot see it otherwise

Answer 13

sorry i meant (b,2) not (b,0)

Answer 14

Okay, these are two equations we get if we plug in points (0,2), (5,4)

(0-a)^2+(2-b)^2
(5-a)^2+(4-b)^2

Since center must be (a,2)
(0-a)^2+(2-2)^2
(5-a)^2+(4-2)^2

If we set them equal to each other
we get
a=29/10

Answer 15

i am willing to admit i am wrong. please show me if it is possible for this not to be so. but if it is true than i think my answer must be correct since the distance between (0,2) and (b,2) is obviously b

Answer 16

So center is (2.9,2)

Answer 17

ok you are using a and i used b. i will use a instead.

Answer 18

yes

Answer 19

i am so lost...

Answer 20

If I am right, you found the center (2.9,2)

Answer 21

ahh ;but i am an idiot because i solved my equation incorrectly. it should be
\[a^2=a^2-10a+25+4\]

Answer 22

and you are right it is a = 2.9

Answer 23

that's the weird answer

Answer 24

no that is a normal answer. i solved a quadratic because i forgot that i was using the square of the distance, not the distance. so i wrote b = something rather than b^2=something

Answer 25

my mistake sorry

Answer 26

So to recap, here is what we did,
1) Y axis is tangent to circle at mid height, so we were able to find part of the center.(a,2)
2) Using two points, we came up with two equation which we solve for a

Answer 27

@perplec if you are still confused:
1) the fact that the circle is tangent at (0,2) must mean the center is at (a,2)
2) the fact that it is a circle means the square of the distance between (0,2) and (a,2) which is a^2 is the same as the square of the distance between (a,2) and (5,4) which is
\[(a-5)^2+(2-4)^2=a^2-10a+29\]

Answer 28

setting
\[a^2=a^2-10a+29\] gives
\[a=\frac{29}{10}=2.9\] and thus the center is
\[(2.9,2)\]

Answer 29

so radius is simply 2.9

Answer 30

right that is the distance, and the equation is
\[(x-2.9)^2+(y-2)^2=2.9^2\]

Answer 31

so the equation of the circle is? plugging the point (2.9,2) and (5,4) and equating them with the distance formula to find the equation of the circle?

Answer 32

oh okay

Answer 33

thanks so much satellite and imranmeah!!!