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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.

Q: 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?

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2

k = 0.1

If the line y = 2x+1.25 is a tangent to the curve y^2 = 10x then it works out that when x = 5/8 then y = 5/2

Tangent to the curve.

You don't.Furthermore, how would you prove your not a virgin? Either way your fu'd.Improved Answer:-y = x+kx2+y2 = 4Substitute y = x+k into the bottom equation:x2+(x+k)(x+k) = 4x2+x2+2kx+k2 = 4k2 = 8 so therefore:2x2+2kx+8-4 = 0 => 2x2+2kx+4 = 0For the straight line to be a tangent to the curve the discriminant b2-4ac of the quadratic equation must equal zero:Hence:-(2*k)2-4*2*4 = 0k2 = 8So: 4*8-4*2*4 = 0 => 32-32 = 0Therefore the straight line y = x+k is a tangent to the curve x2+y2 = 4

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(2, -2)

2

k = 0.1

It is (-0.3, 0.1)

If the line y = 2x+1.25 is a tangent to the curve y^2 = 10x then it works out that when x = 5/8 then y = 5/2

Tangent to the curve.

If: y = kx+1 is a tangent to the curve y^2 = 8x Then k must equal 2 for the discriminant to equal zero when the given equations are merged together to equal zero.

You don't.Furthermore, how would you prove your not a virgin? Either way your fu'd.Improved Answer:-y = x+kx2+y2 = 4Substitute y = x+k into the bottom equation:x2+(x+k)(x+k) = 4x2+x2+2kx+k2 = 4k2 = 8 so therefore:2x2+2kx+8-4 = 0 => 2x2+2kx+4 = 0For the straight line to be a tangent to the curve the discriminant b2-4ac of the quadratic equation must equal zero:Hence:-(2*k)2-4*2*4 = 0k2 = 8So: 4*8-4*2*4 = 0 => 32-32 = 0Therefore the straight line y = x+k is a tangent to the curve x2+y2 = 4

Tangent:In geometry, the tangent line (or simply the tangent) is a curve at a given point and is the straight line that "just touches" the curve at that point. As it passes through the point where the tangent line and the curve meet the tangent line is "going in the same direction" as the curve, and in this sense it is the best straight-line approximation to the curve at that point.Chord:A chord of a curve is a geometric line segment whose endpoints both lie on the outside of the circle.

In geometry a straight line that touches a curve is called a tangent.

A tangent is a line that just touches a curve at a single point and its gradient equals the rate of change of the curve at that point.

A tangent line touches a curve or the circumference of a circle at just one point.