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

Question

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

Bunuel · Accepted Answer

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: https://gmatclub.com/gmat-practice-tests/resources/import/img/m17-12.png Answer: E Tangent.png

cyberjadugar · Answer

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,

cyberjadugar · Answer

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)

Observer · Answer

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.

vipulgoel · Answer

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

sinhap07 · Answer

Hi

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

rahul16singh28 · Answer

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

Crytiocanalyst · Answer

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 ?

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