Thread: Q36. Math - Coordinate

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

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

Originally Posted by clerk

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

Originally Posted by rkandell

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.

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.

C.

I didn't do any sort of paper work. If circle C touches (-1,1) it will also touch (1,1), (-1,1) and (-1,-1) having a center (0,0). Each joining line from (0,0) have a slope of 1 and slope at (-1,1) is parallel to the line from (-1,-1) to (1,1). Everything seems like diagonal of some squares which have slope of 1 each.