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manulath
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Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? 06 Jun 2012, 11:33
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E
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47% (02:19) correct 53% (02:06) wrong based on 316 sessions

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

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

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Bunuel
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Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? 07 Jun 2012, 17:45
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Is the line \(y = kx + b\) tangent to the circle defined by \(x^2 + y^2 = 1\)?

Note that a circle described by the equation \(x^2 + y^2 = 1\) is centered at the origin with a radius of \(r = 1\).

(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 the 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 statements: 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 the 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:




Answer: E

Show Spoiler
Attachment:
Tangent.png
Tangent.png [ 15.97 KiB | Viewed 16742 times ]
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Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? 07 Jun 2012, 01:54
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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
cyberjadugar

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
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Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? Updated on: 07 Jun 2012, 04:58
Originally posted by cyberjadugar on 07 Jun 2012, 00:47.
Last edited by cyberjadugar on 07 Jun 2012, 04:58, edited 1 time in total.
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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.

Hence, answer is (E)
(Note that for some value of b & k the line might be tangent, but using (1) or (2) we can't definitely say so)
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radius.jpg
radius.jpg [ 8.85 KiB | Viewed 14884 times ]

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Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? 07 Jun 2012, 07:32
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manulath
Is line y=kx+b tangent to circle x^2+y^2=1 ?
(1) k+b=1
(2) k^2+b^2=1

OA:
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E
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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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Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? 29 Jun 2015, 18:18
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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
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Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? 26 Sep 2017, 00:08
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Hi

Wont the second stmt imply that we are dealing with a circle and not a line?


Bunuel
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

Answer: E.
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Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? 05 Feb 2018, 21:46
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manulath
Is line y = kx + b tangent to circle x^2 + y^2 = 1 ?

(1) k + b = 1
(2) k^2 + b^2 = 1
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Clearly each statement is Insufficient. Now, combine the two statements. We, have
\(k^2 + b^2 + 2kb = 1\)
\(k*b = 0\). So either k = 0 or b = 0.
When \(k = 0, y =b\) (Here, b can be anything like \(\frac{1}{2}, \frac{-1}{2}, 1, -1\) etc) ---- Not Sufficient.
When \(b = 0, y = kx\) (In this case the line will pass through Origin)

Hence, E
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Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? 26 Jul 2021, 01:15
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The key to solving is getting the X and y intercept correct and the equation of line
Is line y = kx + b tangent to circle x^2 + y^2 = 1

(1) k + b = 1
(kx+b)^2 +x^2 = 1
it can be tangent if (0,1) is satsisfied
it satisfies for b=1
however it might for any other value of b
Therefore insufficient

(2) k^2 + b^2 = 1
let us assume the above case for b=1 it satisfies
however b=0 and k=1
it passes through origin and doesn't hold as tangent

even combined 1 and 2
it doesn't satisfy since the above values can be sustituted
therefore IMO E
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Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? 12 Oct 2023, 02:04
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