Circles and tangent lines

[agresj]

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

1. k + b = 1
2. k^2 + b^2 = 1

[eaakbari]

I would like to know the source of this question , since the level does seem higher than the GMAT. Nevertheless I shall explain.

Stem x^2 + y^2 = 1 is the equation of a circle centered around the origin with radius 1

( Equation of a circle is (x-a)^2 + (y-b)^2 = r^2
where a and b are x and y coordinates of center of circle with radius r


Now to find it there is a tangent to it the tangent should share one common point with the circle. But exactly one point.

Statement one
Tells us nothing pertaining to k b and where it lies. Hence Insuff


Statement two
This statement tells us that the k,b lies on the circle , because of the distance formula which means its 1 away from origin. But this is not necessarily the only point. Hence Insuff


Both combined
Substituting 1 in 2

we get b = 0 or b = 1
which implies k = 1 or k = 0

Which in either case will be a tangent to the circle. Hencer Sufficient

IMO Answer is C

Please give OA and correct me if I am wrong

[agresj]

This is from the GMATClub iPhone app...

Correct answer is E

Statements (1) and (2) combined are insufficient. In equation y = kx + b, k is the slope of the line while b is the elevation (y-coordinate of the point where the line intersects with the vertical axis). Neither statement precludes the line from running through the origin (b = 0, k = 1). In this case, the line intersects circle x^2 + y^2 = 1 at two points and thus is not tangent to it. However, if k = 0 and b = 1, both statements are satisfied but the line only touches the circle at point (0,1). In this case, the line runs parallel to the x-axis.

Is anyone else as confused as I am?

[kstv]

y = kx + b (eq of a st line) tangent to a circle x² + y² = 1
(eq. of a circle with radius = 1)
1. k + b = 1
2. k² + b² = 1
The point of contact will satisfy both the eq.
when the coordinates are (1,1) the y = kx +b will be k+b = 1
cannot think of any other interger coordinates to result in k+b = 1, #
but the point (1,1) cannot lie on the circle x² + y² = 1
so if k+b = 1 the line y = kx +b is not a tangent to the circle
x² + y² = 1.
So is it not sufficient, though it is not affirmative ? How come it is option E.
# but it does not have to be integers infact only at the the points (1,0) (0,1) (-1,0) and (0,-1) will the point of intersection of the tangent and the circle have integer values. At these point y = kx+b will not be k+b = 1

[akhpad]

We should prove that b / sqrt(k^2 + 1) = 1, then only that line to be tangent

Assume b = 0, k = 1
Assume b = 1, k = 0
which certify to both the statement 1 and 2

Statement 1:
Assume b = 0, k = 1 - No
Assume b = 1, k = 0 - Yes
Insufficient

Statement 2:
Assume b = 0, k = 1 - No
Assume b = 1, k = 0 - Yes
Insufficient

Statement 1 and 2
Assume b = 0, k = 1 - No
Assume b = 1, k = 0 - Yes
Insufficient

Ans E

[eaakbari]

I think i got it

When we do combine the 2 statements we substitue 1 in 2

and arrive at
k = 0 then b = 1
if b = 0 then k = 1

k is the slope here, when k = 0 is a horizontal line with y intercept b hence tangent
when k = 1 its a 45 degree line which is will have to pass through origin if b= 0

Hence insuff

Answer E