F.5 Mathematics Question!!!

Method1 :x² + y² - 10 = 0 .....(1)
{
3x + y - 10 = 0 .....(2)By (2) :y = 10 - 3x , sub into (1) :x² + (10 - 3x)² - 10 = 0x² - 6x + 9 = 0(x - 3)² = 0x = 3 (double roots)y = 10 - 3*3 = 1Therefore the straight line 3x+y-10=0 is a tangent to the circle x^2+y^2-10=0
since they only have one touching point (3 , 1).


Method 2 :

The centre of the circle x² + y² - 10 = 0 is (0 , 0)
The radius = √10

The distance between the straight line 3x+y-10=0 and the centre

= |3(0) + 1(0) + (-10)| / √(3² + 1²)

= √10 = the radius.

Therefore the straight line 3x+y-10=0 is a tangent to the circle x^2+y^2-10=0