A circle whose equation is (x-5)^2 + (y+1)^2=36 has a center,C and a radius, r, of

Mathematics

1 answer:

mamaluj [8]2 years ago

3 0

Answer: The center is (5,-1) and radius is 6.

Step-by-step explanation: The equation: (x-h)^2+(y-k)^2=r^2 is really important. Since we have 36 equal r^2, square root it to get 6. And the center would be 5, because (x-5) but do the opposite operation, same for y+1.

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Find the equation of the line that passes through (0, -3) and is parallel to

Tresset [83]

Hey there!

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Answer:

$\green{\boxed{\red{\bold{\sf{y = \dfrac{7}{6}x - 3}}}}}$

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Explanation:

To find the equation of a line, we first have to determine its slope knowing that parallel lines have the same slope.

Let the line that we are trying to determine its equation be $\: \sf{d_1} \:$ and the line that is parallel to $\: \sf{d_1} \:$ be $\: \sf{d_2} \:$ .

$\sf{d_2} \:$ passes through the points (9 , 2) and (3 , -5) which means that we can find its slope using the slope formula:

$\sf{m = \dfrac{\Delta y}{\Delta x} = \dfrac{\green{y_2} - \orange{y_1}}{\red{x_2} - \blue{x_1 }}}$

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⇒Subtitute the values :

$\sf{(\overbrace{\blue{9}}^{\blue{x_1}}\: , \: \overbrace{\orange{2}}^{\orange{y_1}}) \: \: and \: \: (\overbrace{\red{3}}^{\red{x_2}} \: , \: \overbrace{\green{-5}}^{\green{y_2}} )}$

$\implies \sf{m = \dfrac{\Delta y}{\Delta x} = \dfrac{\green{-5} - \orange{2}}{\red{ \: \: 3} - \blue{9 }} = \dfrac{ - 7}{ - 6} = \boxed{ \bold{\dfrac{7}{6} }}}$

$\sf{\bold{The \: slope \: of \: both \: lines \: is \: \dfrac{7}{6}}}$ .

Assuming that we want to get the equation in Slope-Intercept Form, let's substitute m = 7/6:

Slope-Intercept Form:

$\sf{y = mx + b} \\ \sf{Where \: m \: is \: the \: slope \: of \: the \: line \: and \: b \: is \: the \: y-intercept.}$

$\implies \sf{y = \bold{\dfrac{7}{6}}x + b}$ $\\$

We know that the coordinates of the point (0 , -3) verify the equation since it is on the line $\: \sf{d_1} \:$ . Now, replace y with -3 and x with 0:

$\implies \sf{\overbrace{-3}^{y} = \dfrac{7}{8} \times \overbrace{0}^{x} + b} \\ \\ \implies \sf{-3 = 0 + b} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \sf{\boxed{\bold{b = -3}} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$