Q36. Math - Coordinate

Difficulty Level : Medium
Time to Solve : <60 secs

Please discuss your approach and also provide time you took to solve the question.

Q. A circle C is drawn with center as (0,0). A line L is tangent to the circle at point (-1, 1) , what is the slope of L?

A) 0
B) -1
C) 1
D) 1/2
E) None of those



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C ?

Took close to but less than 1 minute. Took a round about way . Looking for alternate methods from others .

Line L is perpendicular to the line joining the center of the circle (0,0) and point at which line L is a tangent. ( -1,1) ( definition of tangent.) Lets call this line R.

Slope of line R = (y2 -y1 ) / (x2-x1) = -1

Therefore slope of line L = ( Slope of line R ) * (-1 )

= 1

C. yeah took less than 60 secs.

the radius of the circle equals (1+1=2) - 2^(0,5)
therefore the equation of the circle x^2+y^2=2,
2x+2y*y'=0, so y'=-x/y
now let put our numbers: y'= -(-1/1)=1
therefore the slope of the tangent equals 1, C

or it's clear that the tangent touches the circle in the middle of circumference of the the circle in the left high quadrant (quart), therefore it has the angle 45, which is tag45=1

C - 20 secs

Found the slope of the radius of the circle that touches the tangent point (-1,1). m = 1/-1 = -1. Since the line is perpendicular to the radius's slope the slope of the line is the opposite of the reciprocal. Therefore (-1/1)*-1 = 1.

C

Here is much easier way. The tangent to the circumference L will cross X axis at -1+x and also will cross Y axis at 1+y. Where -1+x=1+y. This tells us that we have an isosceles right triangle with angles 45,45,90. Thus tan45=1.

Dedicated to Clerk. Think easily

Good solutions everyone! Its amazing to see so many different solutions.

For those who dont know trigonometry , this solution by rkandell can be used as a quick method to solve the question.

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