Write the equation for the line tangent to the circle x2 + y2 = 29 at the point (2, 5).

1 answer:

3 0

X2 + y2 = 29

Center of the circle
x=0 and y=0

Therefore the slope of the line perpendicular to the tangent line
(y2-y1) / (x2-x1)
(5-0) / (2-0)
5/2

so the slope of the tangent
-2/5

General equation for the line
y = mx + c
if the line satisfies with the point (2,5)
5 = 2m + c

m = -2/5

5 = 2 * (-2/5) + c
c = 5 + (4/5)
c = 29/5

Therefore the equation of the line
y = mx + c
y = (-2/5) m + 29/5

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The polynomial of degree 5, P ( x ) has leading coefficient a=1, has roots of multiplicity 2 at x = 3 and x = 0 , and a root of

ser-zykov [4K]

Answer:

$p_{5} (t) = x^{5} - 5\cdot x^{4} + 3\cdot x^{3} +9\cdot x^{2}$ , for $r_{1} = 0$

Step-by-step explanation:

The general form of quintic-order polynomial is:

$p_{5}(t) = a\cdot x^{5} + b\cdot x^{4} + c\cdot x^{3} + d\cdot x^{2} + e \cdot x + f$

According to the statement of the problem, the polynomial has the following roots:

$p_{5} (t) = (x - r_{1})\cdot (x-3)^{2}\cdot x^{2} \cdot (x+1)$

Then, some algebraic handling is done to expand the polynomial:

$p_{5} (t) = (x - r_{1}) \cdot (x^{3}-6\cdot x^{2}+9\cdot x) \cdot (x+1)\\p_{5} (t) = (x - r_{1}) \cdot (x^{4}-5\cdot x^{3} + 3 \cdot x^{2} + 9 \cdot x)$