Find the points of intersection of the line $$4x-3y=10$$​ and the circle $$x^2+y^2-2x+4y-20=0$$​.

Correct option is A

Given:
We need to find the points of intersection between the line 4x - 3y = 10 and the circle $$x^2 + y^2 - 2x + 4y - 20 = 0 .$$​
Solution:

Option (a) (4, 2) and (-2, -6):
Put (4, 2) into both equations:
Line: 4(4) - 3(2) = 16 - 6 = 10 
Circle: $$4^2 + 2^2 - 2(4) + 4(2) - 20 = 16 + 4 - 8 + 8 - 20 = 0$$​​

Put (-2, -6):
Line: 4(-2) - 3(-6) = -8 + 18 = 10 
Circle:$$(-2)^2 + (-6)^2 - 2(-2) + 4(-6) - 20 = 4 + 36 + 4 - 24 - 20 = 0$$​​
Both points satisfy the original equations.

Option (b) (-2, 3) and (4, 2):

Line: 4(-2) - 3(3) = -8 - 9 = -17$$\not=10$$

Option (b) is wrong.

Option (c) (-2, 3) and (4, 3):

Line: 4(-2) - 3(3) = -8 - 9 = -17$$\not=10$$

​Option (c) is wrong.

Option (d) (2, -3) and (4, 2):

Line: 4(2) - 3(-3) = 8 + 9 = 17$$\not=10$$

Option (d) is wrong.

Thus, option (a) is right.