The number of circles that can be drawn out of $10$ points of which $7$ are co-linear

Question

My attempt

The number of cases should be: $$10C2+3C2\cdot7C2+3C2\cdot7C1+3C3\cdot7C1+3C3\cdot7C2$$

But the answer is $85$. How?

Answer 1

3 non-colinear points are needed to draw a circle.

GIVEN: 10 points are given of which 7 points are colinear.

There are 3 ways this can be done

1) choosing three remaining points that are non co-linear = 3C3 = 1

2) choosing 2 points from the 3 remaining points and 1 point from the 7 colinear points = 3C2 * 7C1 = 3 * 7 = 21

3) choosing 1 point from the 3 remaining points and 2 points from the 7 colinear points = 3C1 * 7C2 = 3 * 21 = 63

Total = 63 + 21 + 1 = 85

Answer 2

There are 3 ways that 2 of the non-collinear points can combine with one of the collinear points. There are 2 from 7 ways (21) that 2 of the collinear points can combine with one of the 3 non-collinear points. There is one additional way for the 3 non-collinear points.

$7(3C2) + (7C2)3 + 1$

$(7*3) + (21*3) + 1= 85$