Current location - Training Enrollment Network - Mathematics courses - A math problem in junior high school circle. .
A math problem in junior high school circle. .
Solution: The analysis is as follows

Connect AC and MC and make a straight line DE tangent to CM. Because MA equals 2 and OM equals 1, OA equals 1.

So OA=OM, because OC=OC, angle AOC= angle MOC = 90, so triangle AOC is all equal to triangle MOC.

So AC=MC, because MC=AM and triangle ACM is an equilateral triangle, so angle CMA is 60, because DC is perpendicular to CM, so angle DCM is 30, so CM is half of DM, so DM is equal to 4, so the coordinate of D is (-3,0), because MB is equal to 2 and OM is equal to 1, so B (3 3,0).

Bringing C and D into a linear function, it is found that K equals 2, B equals 6, and the analytical formula is Y=2X+6?

There may be some simple steps, so please don't pay attention.