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

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Kathleen Mcpherson · Accepted Answer

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+2y⋅dydx=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:  (y−y1)=m(x−x1)  The tangent would therefore have the equation:  (y−12)=−512(x−5)  12y-144=-5x+25  12y=-5x+169  y=112(−5x+169)

surgescasjag · Accepted Answer

Explanation:Differentiate implicitly:ddx(x2+y2)=02x+2ydydx=0dydx=−xySo, for x=5, y=12:y′(5)=[dydx]5,12=−512The equation of the tangent line is:y=y0+y′(x0)(x−x0)=12−512(x−5)y=−512x+144+251212y+5x=169

How do you write an equation of the line tangent

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