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3 years ago

A circle passes through (3,-1),(-2, 4), and (6,8)

Mathematics

1 answer:

Vaselesa [24]3 years ago

5 0

Answer:

D = -6, E = -8 , F = 0

Step-by-step explanation:

standard form = $x^{2} + y^{2} + Dx + Ey + F = 0$

now, when circle passes through (3,-1);

⇒ $3^{2} + (-1)^{2} + D(3) + E(-1) + F = 0$

⇒ $3D - E + F + 10 = 0$ ...............( equation 1)

when circle passes through (-2,4);

⇒ $(-2)^{2} + 4^{2} + D(-2) + E(4)+ F = 0$

⇒ $-2D + 4E + F + 20 = 0$ ...............( equation 2)

when circle passes through (6,8);

⇒ $6^{2} + 8^{2} + D(6) + E(8) + F = 0$

⇒ $6D + 8E + F + 100= 0$ ................( equation 3)

by solving these 3 equations , we get;

D = -6, E = -8, F = 0

hence,

standard form = $x^{2} + y^{2} + Dx + Ey + F = 0$

= $x^{2} + y^{2} - 6x - 8y = 0$

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3 years ago

Show that if the vector field F = Pi + Qj + Rk is conservative and P, Q, R have continuous first-order partial derivatives, then

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Step-by-step explanation:

Given vector field,

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Where,

$P=f_x=\frac{\partial f}{\partial x}, Q=f_y=\frac{\partial f}{\partial y}, R=f_z=\frac{\partial f}{\partial z}$

To show,

$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}, \frac{\partial P}{\partial z}=\frac{\partial R}{\partial x}, \frac{\partial Q}{\partial z}=\frac{\partial R}{\partial y}$

Consider,

$\frac{\partial P}{\partial y}=\frac{\partial}{\partial y}(\frac{\partial f}{\partial x})=\frac{\partial^2 f}{\partial y\partial x}=\frac{\partial^2 f}{\partial x\partial y}=\frac{\partial }{\partial x}(\frac{\partial f}{\partial y})=\frac{\partial Q}{\partial x}$

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$\frac{\partial Q}{\partial z}=\frac{\partial}{\partial z}(\frac{\partial f}{\partial y})=\frac{\partial^2 f}{\partial z\partial y}=\frac{\partial^2 f}{\partial y\partial z}=\frac{\partial}{\partial y}(\frac{\partial f}{\partial z})=\frac{\partial R}{\partial y}$

Hence proved.

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