Question

Prove that the radius of the circle with equation x² + y² = 25 is perpendicular on the tangent line in P(3,4) to the circle? Thank you!

Answer 1

The slope of the Tangent line is `-3/4` and the slope of the Radius is `4/3`, The slopes are negative reciprocal of each other. Therefore they are perpendicular to each other.

Explanation:

Use the equations and slopes of the Radius and the Tangent line.

For the Radius whose end points are the center `(0, 0)` and `(3, 4)`
The slope is `4/3`

For the Tangent line at `(3, 4)`

Determine the slope by differentiating `x^2+y^2=25`
Let `y=sqrt(25 - x^2)`

`dy/dx=1/(2sqrt(25-x^2))*-2x`

`dy/dx=(-x)/(sqrt(25-x^2))`

By the given point `(3, 4)` use `x=3` in `dy/dx`
`dy/dx=(-x)/(sqrt(25-x^2))`

`dy/dx=(-3)/(sqrt(25-(3)^2))`

`dy/dx=-3/4`

The slope of the Tangent line is `-3/4` and the slope of the Radius is `4/3`, The slopes are negative reciprocal of each other. Therefore they are perpendicular to each other.

I hope the explanation is useful...God bless.

Answer 2

See below

Explanation:

Given `x^2+y^2=25`

We know this is a circle centered in `(0,0)`

The given point belongs to circle because

`3^2+4^2=25`

The derivative is

`2x+2yy´=0` valuated in `P` is

`6+8y´=0` this give us `y´=-6/8=-3/4` this the slope of tangent line to circle in `P`

By other hand the vector defined by `(3,4)` form an angle with x-axis which tangent is `4/3`

We know that two vectors are penperdiclar if the product of tangents of their angles is `-1`

But `-3/4·4/3=-1`. So radius and tangent are perpendicular