How do you write an equation of the line tangent

Sarah-Louise Prince

Sarah-Louise Prince

Answered question

2022-02-15

How do you write an equation of the line tangent to x2+y2=169 at the point (5,12)?

Answer & Explanation

Kathleen Mcpherson

Kathleen Mcpherson

Beginner2022-02-16Added 9 answers

Here we have the equation of a circle: x2+y2=132
To determine the slope of a tangent to the circle at any point we need to use implicit differentiation.
x2+y2=132
2x+2ydydx=0
dydx=xy
At the point (5,12), dydx=512
So the tangent has a slope of 512 and passes through the point (5,12)
The equation of a line of slope m, passing through the point (x1,y1) is:
(yy1)=m(xx1)
The tangent would therefore have the equation:
(y12)=512(x5)
12y-144=-5x+25
12y=-5x+169
y=112(5x+169)
surgescasjag

surgescasjag

Beginner2022-02-17Added 10 answers

Explanation:
Differentiate implicitly:
ddx(x2+y2)=0
2x+2ydydx=0
dydx=xy
So, for x=5, y=12:
y(5)=[dydx]5,12=512
The equation of the tangent line is:
y=y0+y(x0)(xx0)=12512(x5)
y=512x+144+2512
12y+5x=169

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?