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Re: Line y = kx + b tangent to circle [#permalink]
06 Jun 2012, 23:47

1

This post received 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 circlex^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)

Re: Line y = kx + b tangent to circle [#permalink]
07 Jun 2012, 00:41

cyberjadugar 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

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

2

This post received 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.

Regards,

manulath wrote:

cyberjadugar 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

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

1

This post received KUDOS

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 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.

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.

Re: Is line y=kx+b tangent to circle x^2+y^2=1 ? [#permalink]
07 Jun 2012, 16:45

1

This post received KUDOS

Expert's post

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: