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Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? [#permalink]
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06 Jun 2012, 11:33
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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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Re: Line y = kx + b tangent to circle [#permalink]
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06 Jun 2012, 11:35
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]
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07 Jun 2012, 00:47
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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 ykxb=0 to be tangesnt, the shortest distance between line & (0,0) = radius = 1 or \((0k*0b)/\sqrt{1+k^2}=1\) or \(b=\sqrt{1+k^2}\) squaring both sides, \(b^2=1+k^2\) so, question can be reframed as Is \(b^2k^2=1\)? 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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Last edited by cyberjadugar on 07 Jun 2012, 04:58, edited 1 time in total.



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Re: Line y = kx + b tangent to circle [#permalink]
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07 Jun 2012, 01:41
cyberjadugar wrote: For line ykxb=0 to be tangesnt, the shortest distance between line & (0,0) = radius = 1 or \((0k*0b)/\sqrt{1+k^2}=1\) 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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Re: Line y = kx + b tangent to circle [#permalink]
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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 ymxc=0 then the distance \(d = (y_1mx_1c)/\sqrt{(coff. of. y)^2+(coff. of. x)^2}\) thus, d = \((y_1mx_1c)/\sqrt{(1)^2+(m)^2}\) I could not think of any other direct approach for now. Regards, manulath wrote: cyberjadugar wrote: For line ykxb=0 to be tangesnt, the shortest distance between line & (0,0) = radius = 1 or \((0k*0b)/\sqrt{1+k^2}=1\) 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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Re: Line y = kx + b tangent to circle [#permalink]
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07 Jun 2012, 02:25
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cyberjadugar wrote: 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 ymxc=0 then the distance \(d = (y_1mx_1c)/\sqrt{(coff. of. y)^2+(coff. of. x)^2}\) thus, d = \((y_1mx_1c)/\sqrt{(1)^2+(m)^2}\) 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]
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07 Jun 2012, 07:32
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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: 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=1k b^2=1k^2 => b^2=(1k)^2=12k+k^2=1k^2 ==> 2k^22k=0 => k(2k2)=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]
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07 Jun 2012, 17:45
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Is line \(y=kx+b\) tangent to circle \(x^2+y^2=1\)?(1) \(k+b=1\) (2) \(k^2+b^2=1\) 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 4428 times ]
Answer: E.
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Re: Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? [#permalink]
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Re: Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? [#permalink]
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29 Jun 2015, 18:18
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^21)=0 for real roots D must be >/ 0 hence + 4k^2b^2 4(k^2+1)(b^21) >/ 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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Re: Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? [#permalink]
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Re: Is line y = kx + b tangent to circle x^2 + y^2 = 1 ?
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