equation of a tangent line

Question

write the equation of the line that is tangent to x^2+y^2=100; (8,6)

Answer 1

Line equation is y = mx+b

x^2+y^2 = 100

Compare it a circle equation (x-h)^2+(y-k)^2 = r^2

The circle through the origin.

The points say (x1,y1) =(0,0)

(x2,y2) = (8,6)

Slope m1 = (6-0)/(8-0)

= 6/8

= 3/4

Tangent line slope say m2 .

We know that m1*m2 = -1

3/4m2 = -1

Multiply to each side by 4/3.

3/4*4/3 m2 = -1*4/3

m2 = -4/3

Equation to tangent line is y-y1 = m2(x-x1) and point (8,6)

y-6 = -4/3(x-8)

3(y-6)  = -4(x-8)

3y-18 = -4x+32

Add 18 to each side.

3y-18+18 = -4x+32+18

3y = -4x+50

Divide to each side by 3.

3y/3 = (-4x+50)/3

y = (-4/3)x+50/3

Required line equation is y  = (-4/3)x+50/3.

Answer 2

The equation is x ² + y ² = 100 and the point is (8, 6).

We can use implicit differentiation to find y '.

The implicit(or total) derivation of x ² + y ² = 100 is,

2x + 2yy ' = 0

x + yy ' = 0

yy ' = - x

y ' = - x /y

Substitute the values of (x, y ) = (8, 6) in the above equation.

y ' = - 8/6

y ' = - 4/3.

This is the slope (m ) of the tangent line to x ² + y ² = 100 at (8, 6).

To find the tangent line equation, substitute the values of m = - 4/3 and (x, y ) = (8, 6) in the slope - intercept form of an equation : y = mx + b.

6 = (- 4/3)(8) + b

b = 6 + (32/3)

b = ( 18 + 32)/3

b = 50/3.

Substitute m = - 4/3 and b = 50/3 in y = mx + b.

y = (- 4 / 3)x + 50/3.

The tangent line equation is y = (- 4 / 3)x + 50/3.