Find the equation of the tangent line to the equation y=x^2 at x=1. Then, this line is also tangent to a circle with center (8, 0) at x = 2. Find the equation of this circle.

I got y=2x+1 for my tangent line (which is wrong), and had no idea how to get the second part.

Question

Find the equation of the tangent line to the equation y=x^2 at x=1. Then, this line is also tangent to a circle with center (8, 0) at x = 2. Find the equation of this circle. 

I got y=2x+1 for my tangent line (which is wrong), and had no idea how to get the second part.

anonymous · Answer

So when they say find tangent line to y=x^2 that means take derivative of that first.

anonymous · Answer

Do you know derivative of x^2 ?

anonymous · Answer

yes, it is 2x. and then plug in x=1 and you get 2. therefore the slope is 2 and then plug x=1 back in to the original equation and get 1. therefore y=2x+1.....that was how I got it at least

anonymous · Answer

yet when I submitted that for my answer, it said it was incorrect...

campbell_st · Answer

so next find the perpedicular distance from a point to a line

the line is y = 2x + 1   the point is (8, 0) 

this will give the radius of the circle. Also the tangent to a circle is perpendicular to the radius. 

then the standard form is 

$$(x - h)^2 + (y - k)^2 = r^2$$

(h, k) is the centre and r is the radius

anonymous · Answer

y=2x-1 should be the correct tangent equation.

anonymous · Answer

why is it -? the equation is y=mx+b and there isn't a negative value to make it negative...

anonymous · Answer

Graph x^2. at x=1 the tangent line is represented by 2x. So if you think of rise of run. Imagine going to the left from x=1 on the graph, and you should go down two  units.

campbell_st · Answer

the y intercept is -1 

the point on the parabola is (1, 1) 

the general from is 

y = mx + b

so 

y = 2x + b    substitute the point x = 1 and y = 1 to find b...

anonymous · Answer

So if the y intercept is -1 wouldn't the equation be y=2x-1?

campbell_st · Answer

that's correct...

campbell_st · Answer

so you need to now find the perpendicular distance from a line to a point to get the radius of the circle.

anonymous · Answer

ok, I get that now...now what about the second part. I know campbell_st started it, but is there anything else I should know before I try to do it?

anonymous · Answer

Just use the distance formula between the two points.

campbell_st · Answer

nope the tangent and radius of the circle are perpendicular to each other. 

so find the radius, you know the centre so substitute it into the standard form of the equation.

anonymous · Answer

ok, and just to check, the radius is 6 since the center is at (8,0) and and they want to know for the point when x=2 (8-2=6)

campbell_st · Answer

so if you have a line $$ax + by + c = 0$$

the perpendicular distance is 

$$d = \frac{\left| ax_{1} + by_{1} + c \right|}{\sqrt{a^2 + b^2}}$$

so you have

$$d = \frac{\left| 2\times 8 - 1\times0 -1 \right|}{\sqrt{2^2 + 1^2}}$$

anonymous · Answer

assuming that is true, so far, for my equation, I have 36+y^x=36 which does not seem right

anonymous · Answer

I have no clue what you just typed and have no idea what I am supposed to do with that...

campbell_st · Answer

well a slight correction, at x = 2 is a point on the tangent line and the circle's circumference... a point of intersection.... 

so if the tangent is y = 2x - 1   when x = 2  the point is y = 2*2 - 1 

so the point is (2, 3)

anonymous · Answer

I am honestly so lost. Can someone just do the work and show me because I have no clue what to do at all

campbell_st · Answer

ok... it's a formula for finding the perpendicular distance from a point to a line

anonymous · Answer

ok, but I honestly need you to tell me what to do because I am lost

campbell_st · Answer

ok... so you have 2 points (2, 3) and (8, 0) find the distance between them.

this will be the radius


and I've attached a graph

campbell_st · Answer

so leave the radius as a radical 

and then the standard form of a circle is 

$$(x - h)^2 + (y - k)^2 = r^2$$

you know the centre h = 8, k = 0

and you calculate the radius... 

so substitute them

anonymous · Answer

what is the distance formula again?

anonymous · Answer

nevermind, I looked it up...the distance is root(45)

campbell_st · Answer

given 2 points 

$$(x_{1}, y_{1})~~~(x_{2},y_{2})$$

the distance between them is 

$$d = \sqrt{x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$$

or just use pythagoras' theorem 

8 to 2 = 6  and up 3  

so d^2 = 6^2 + 3^2

anonymous · Answer

i know this goes against what the purpose of this site is, and I do appreciate all of your help, but my homework is due soon, midnight my time, as it is online homework, can you please give me the answer as I still have other problems I need to do too. I have tried to do them while working on this one, but this has taken too much to focus on the others also. I am sorry, but I really need the answer. I will be willing to work it out later though

campbell_st · Answer

ok... so here is the guts of it 

|dw:1413874315525:dw|

just find r^2. 

the circle is and substitute your value for r^2

$$(x - 8)^2 + y^2 = r^2$$

anonymous · Answer

r^2=45

campbell_st · Answer

great... now substitute it for the equation of the circle

anonymous · Answer

(x-8)+y^2=45 Thanks!

campbell_st · Answer

its (x - 8)^2

anonymous · Answer

I meant to type that...whoops!