Line M is tangent to a circle, which is centered on point

Hi Anaira,

Let me discuss two ways to approach this question.

1. The first (the approach that we personally endorse) is to plot the question on the XY plane. Plotting any coordinate geometry question visually helps us solve the question and bypasses the use of ugly algebraic equations.

We know that the line M is the tangent to the circle with center say C (3,4). This is a Yes/No DS question where we need to see if line M runs through the point (6,6).

Statement 1 : Line M runs through point (-8, 6)

This statement does not give us any information about the radius of the circle. From any external point we can draw two tangents to a circle. So the line M can pass through the point (6,6) and be a tangent at (3,6) or the line M can be a tangent to the circle without passing through the point (6,6). Please refer to the figure below. Insufficient



Statement 2 : Line M is tangent to the circle at point (3, 6)

Since line M is tangent at (3,6), the radius has to be 2 units. We know that a tangent is always perpendicular to the radius at the point of contact, the only tangent possible here will be the line y=6 and this line will always pass through the point (6,6). Refer to the diagram given below. Sufficient



OA : B

2. The second is the algebraic approach which uses the concept of a slope of a line. It is very clear that if the two statements are taken together than we can easily find the slope of the line (0 in this case) and the y intercept here will be 6, so the equations of the line will be y = 6, but this question is a classic C trap, so we need to evaluate the statements individually first before picking option C.

Statement 1 : Just gives us one point (-8,6), to find the slope we require a minimum of two points. Insufficient

Statement 2 : We can find the slope of the line using the points (3,4),the center of the circle, and the point (3,6) the point of tangency. The slope here will be undefined i.e. (6-4)/(3-3) which means that the line is parallel to the y axis. Since we know that the radius and tangent are perpendicular, the slope of the the line M must be parallel to the X axis. The slope of a line parallel to the X axis is 0 and the equation of this line will be y = c. The y intercept c here will be 6 making the equation of line M y = 6. Coordinate (6.6) will definitely lie on the line y = 6. Sufficient.

CrackVerbal Academics Team
_________________

From an external point, you can draw 2 tangents to a given circle. Moreover, we have no information about the radius of the circle. Thus, for different radii, we would have two unique sets of tangents and we cannot deduce if any of them distinctly passes through the point (6,6).

From F.S 2, the point of tangency is (3,6) and thus the radius has to be 2 units. Also, as a tangent is perpendicular to the radius(which is represented by the line x = 3) at the point of tangency, the only tangent possible is the line y=6. Sufficient.

_________________

Can any one explain this problem?

Is there any algebraic or conceptual way to do above problem?
I don't think geometrical approach will be able to solve this within 2 minutes.

Thank You for your valuable inputs on this problem. It's clear to me now and seems geometrical approach is far much efficient than algebraic one.

statement 1 is clearly insuff.

statement 2- line drawn to the tangent from center of a circle is perpendicular to tangent.we know the point touching the circle(tangent coordinate) and radius coordinates , we can get slope of the tangent using .m1*m2= -1(as lines are perpendicular)
we can deduce the equation for the tangent .
finally we can test for point 6,6
suff.

When i tried to solve the problem, i got D. is the answer D?

The OA (Official Answer) is given under the spoiler in the original post, and it's B.
_________________

This question can be solved even without using pen and paper under 1 minute - by the below mentioned trick -
Only option B provides the radius(indirectly) of the circle hence the equation of the tangent can be calculated.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________