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# Is line y = kx + b tangent to circle x^2 + y^2 = 1 ?

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Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? [#permalink]  06 Jun 2012, 10:33
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42% (02:36) correct 58% (01:20) wrong based on 62 sessions
Is line y = kx + b tangent to circle x^2 + y^2 = 1 ?

(1) k + b = 1
(2) k^2 + b^2 = 1
[Reveal] Spoiler: OA
Manager
Joined: 12 May 2012
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Kudos [?]: 62 [0], given: 14

Re: Line y = kx + b tangent to circle [#permalink]  06 Jun 2012, 10:35
manulath wrote:
Is line y=kx+b tangent to circle x^2+y^2=1 ?
(1) k+b=1
(2) k^2+b^2=1

I was unable to grasp the explanation. Kindly help me with detailed explanation
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Re: Line y = kx + b tangent to circle [#permalink]  06 Jun 2012, 23:47
1
KUDOS
Hi,

To check the tangency of line to a circle:
if the shortest distance from the center of the circle to the line is equal to radius then the line is tangent to circle.

center of circle$$x^2+y^2=1$$ is (0,0)

For line y-kx-b=0 to be tangesnt, the shortest distance between line & (0,0) = radius = 1
or $$|(0-k*0-b)/\sqrt{1+k^2}|=1$$
or $$|b|=\sqrt{1+k^2}$$
squaring both sides,
$$b^2=1+k^2$$
so, question can be reframed as Is $$b^2-k^2=1$$?

Clearly, neither (1) nor (2) proves the above relation.

(Note that for some value of b & k the line might be tangent, but using (1) or (2) we can't definitely say so)
Attachments

radius.jpg [ 8.85 KiB | Viewed 3143 times ]

Last edited by cyberjadugar on 07 Jun 2012, 03:58, edited 1 time in total.
Manager
Joined: 12 May 2012
Posts: 83
Location: India
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GMAT 1: 650 Q51 V25
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Kudos [?]: 62 [0], given: 14

Re: Line y = kx + b tangent to circle [#permalink]  07 Jun 2012, 00:41
For line y-kx-b=0 to be tangesnt, the shortest distance between line & (0,0) = radius = 1
or $$|(0-k*0-b)/\sqrt{1+k^2}|=1$$
or $$|b|=\sqrt{1+k^2}$$
squaring both sides,
$$b^2=1+k^2$$

It would be really helpful if you can explain the above.
How that particular equation was arrived at?
Thanks
Senior Manager
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Kudos [?]: 287 [2] , given: 23

Re: Line y = kx + b tangent to circle [#permalink]  07 Jun 2012, 00:54
2
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Hi,

This is the formula from 10+ maths,

to find the shortest distance of a point $$(x_1,y_1)$$ from line y=mx+c
change the equation of line to y-mx-c=0
then the distance $$d = |(y_1-mx_1-c)/\sqrt{(coff. of. y)^2+(coff. of. x)^2}|$$
thus, d = $$|(y_1-mx_1-c)/\sqrt{(1)^2+(-m)^2}|$$
I could not think of any other direct approach for now.

Regards,
manulath wrote:
For line y-kx-b=0 to be tangesnt, the shortest distance between line & (0,0) = radius = 1
or $$|(0-k*0-b)/\sqrt{1+k^2}|=1$$
or $$|b|=\sqrt{1+k^2}$$
squaring both sides,
$$b^2=1+k^2$$

It would be really helpful if you can explain the above.
How that particular equation was arrived at?
Thanks
Manager
Joined: 12 May 2012
Posts: 83
Location: India
Concentration: General Management, Operations
GMAT 1: 650 Q51 V25
GMAT 2: 730 Q50 V38
GPA: 4
WE: General Management (Transportation)
Followers: 2

Kudos [?]: 62 [1] , given: 14

Re: Line y = kx + b tangent to circle [#permalink]  07 Jun 2012, 01:25
1
KUDOS
Hi,

This is the formula from 10+ maths,

to find the shortest distance of a point $$(x_1,y_1)$$ from line y=mx+c
change the equation of line to y-mx-c=0
then the distance $$d = |(y_1-mx_1-c)/\sqrt{(coff. of. y)^2+(coff. of. x)^2}|$$
thus, d = $$|(y_1-mx_1-c)/\sqrt{(1)^2+(-m)^2}|$$
I could not think of any other direct approach for now.

This is a sure shot winner. A helpful formula and time saver.
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Re: Line y = kx + b tangent to circle [#permalink]  07 Jun 2012, 06:32
1
KUDOS
manulath wrote:
Is line y=kx+b tangent to circle x^2+y^2=1 ?
(1) k+b=1
(2) k^2+b^2=1

OA:
[Reveal] Spoiler:
E

Although the other solutions are mathematically correct, I think there's an easier way than memorizing a formula.

We need to solve for k and b.

1. A single equation does not allow us to solve for two variables. Additionally, k=1, b=0 gives an answer of no and k=0, b=1 gives an answer of yes. Insufficient.
2. See 1. Insufficient.

1+2. Taken together we have:
b=1-k
b^2=1-k^2
=> b^2=(1-k)^2=1-2k+k^2=1-k^2
==> 2k^2-2k=0 => k(2k-2)=0, and therefore k=0 or k=1.
In the case of k=0, b=1. Then the equation for the line is y=1, and the answer is YES. In the case of k=1, b=0. Then the equation for the line is y=x, and the answer is NO. Insufficient.

E.
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Re: Is line y=kx+b tangent to circle x^2+y^2=1 ? [#permalink]  07 Jun 2012, 16:45
2
KUDOS
Expert's post
1
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BOOKMARKED
Is line $$y=kx+b$$ tangent to circle $$x^2+y^2=1$$?

(1) $$k+b=1$$
(2) $$k^2+b^2=1$$

Notice that a circle represented by the equation $$x^2+y^2=1$$ is centered at the origin and has the radius of $$r=\sqrt{1}=1$$.

(1) $$k+b=1$$ --> if $$k=0$$ and $$b=1$$ then the equation of the line becomes $$y=1$$ and this line is tangent to the circle but if $$k=1$$ and $$b=0$$ then th equation of the line becomes $$y=x$$ and this line is NOT tangent to the circle. Not sufficient.

(2) $$k^2+b^2=1$$. The same example is valid for this statement too. Not sufficient.

(1)+(2) Again the same example satisfies both statement: if $$k=0$$ and $$b=1$$ then the equation of the line becomes $$y=1$$ and this line is tangent to the circle but if $$k=1$$ and $$b=0$$ then th equation of the line becomes $$y=x$$ and this line is NOT tangent to the circle. Not sufficient. Look at the diagram below to see both cases:
Attachment:

Tangent.png [ 15.97 KiB | Viewed 3334 times ]

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Re: Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? [#permalink]  06 Feb 2015, 07:43
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Re: Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? [#permalink]  29 Jun 2015, 17:18
Alt way,

For line to be tangent at circle , by putting equation of line in equation of circle we should get real roots, I mean the D must be >/ 0

x^2 + (Kx+b)^2 =1
x^2(k^2+1) + 2kbx +(b^2-1)=0
for real roots D must be >/ 0
hence
+- 4k^2b^2 -4(k^2+1)(b^2-1) >/ 0

from either statements we cant find whether above statement is true or not , hence E

Experts please let me know if my reasoning is correct
Re: Is line y = kx + b tangent to circle x^2 + y^2 = 1 ?   [#permalink] 29 Jun 2015, 17:18
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