How do you write an equation of the line tangent to (x−1)2+(y−1)2=25 at the point (4,-3)?

Question

Lillie-May Sutton · Accepted Answer

Expand the equation of the circle.  x2−2x+1+y2−2y+1=25  x2+y2−2x−2y+2=25  Differentiate both sides with respect to x using implicit differentiation and the power rule.  ddx(x2+y2−2x−2y+2)=ddx(25)  2x+2y(dydx)−2−2(dydx)=0  2y(dydx)−2(dydx)=2−2x  dydx(2y−2)=2−2x  dydx=2−2x2y−2  dydx=2(1−x)2(y−1)  dydx=1−xy−1  dydx=−x−1y−1  The slope of the tangent is given by evaluating f(x,y) inside the derivative.  mtan≥nt=−4−1−3−1  mtan≥nt=−3−4  mtan≥nt=34  The equation of the tangent is therefore:  y−y1=m(x−x1)  y−(−3)=34(x−4)  y+3=34x−3  y=34x−6

Pregazzix2a · Accepted Answer

The equation is that of a circle centred at point O(1,1), Let P be the point (4, -3).
The tangent at P is at right angles to the radius at P. But the gradient of radius OP
is

\frac{- 3 - 1}{4 - 1} = - \frac{4}{3}

. Therefore the gradient of the tangent is

\frac{- 1}{- \frac{4}{3}} = + \frac{3}{4}

.
Therefore the equation of the tangent is

y = \frac{3}{4} x + c

for some c. Since P lies on the tangent,

- 3 = \frac{3}{4} 4 + c

. Hence c=-6.

How do you write an equation of the line tangent

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