How do you prove that the straight line y equals x plus c is a tangent to the curve x squared plus y squared equals 4 when c squared equals 8?

Answer 1

The general form of a line tangent to a circle is:y=mx+a(1+m2 )1/2. where "a" is the radius of the circle. Here circle is x2 + y2=4, so radius=a=2. now

c2=a2(1+m2)=8 (given)

or 8=4(1+m2)

2=1+m2 or

m2=1 or

m=1. so equation becomes

y=mx+c or

y=x+c

Improved Answer:

Equation 1: y = x+square root of 8 => y2 = x2+square root of 32x+8 when both sides are squared.

Equation 2: y2 = 4-x2

By definition:

x2+the square root of 32x+8 = 4-x2 => 2x2+the square root of 32x+4 = 0

If the discriminant b2-4ac of the above quadratic equation is equal to zero then this is proof that the straight line is tangent to the curve:

b2-4ac = the square root of 322-4*2*4 = 0

Therefore the straight line is a tangent to the curve because the discriminant of the quadratic equation equals zero.