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2 years ago

PLEASE HELP QUICKLY

Mathematics

1 answer:

iren [92.7K]2 years ago

7 0

Answer:

The exponents of a start at n and decrease until they reach 0 . The exponents of b start at 0 and increase until they reach n

Step-by-step explanation:

The general formula for $(a+b)^n$ is:

$a^n + C_1a^{n-1}b + C_2a^{n-2}b^2 + ... + C_{n-1}a^2b^{n-2} + C_nab^{n-1} + b^n$

Therefore, the exponents of a start at n and decrease until they reach 0 . The exponents of b start at 0 and increase until they reach n

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Find the equation of the circle with a diameter whose end points are (-1,-2) and (-3,2)

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The equation of the circle with a diameter whose end points are (-1,-2) and (-3,2) is

$x^{2} +y^{2}+4x-1=0$

Step-by-step explanation:

Explanation:-

Step 1:-

The equation of the circle having center and radius is

$(x-h)^2+(y-k)^2=r^2$

here center is (h,k) and radius is r

Given diameter whose end points are (-1,-2) and (-3,2)

The diameter of the circle is passing through the center of the circle

so center of the circle = midpoint of two end points

$(\frac{-1 +(-3) }{2} ,\frac{-2+2 }{2} )$

$(-2,0)$

therefore center (h,k) = (-2,0)

Step 2:-

we have to find the radius of the circle

the radius of the circle = the distance from center to the one end point

i.e., C P = r

Given one end point is P(-3,2) and center C(-2,0)

The distance formula of two points are

$\sqrt{(x_{2}-x_{1} ) ^{2}+ (y_{2}-y_{1} ) ^{2}}$

$r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}$

$r=\sqrt{5}$

Step 3:-

center (h,k) = (-2,0) and

radius $r=\sqrt{5}$

The standard form of circle equation

$(x-h)^2+(y-k)^2=r^2$

$(x-(-2))^2+(y-0)^2=\sqrt{5} ^2$

on simplification is

$x^{2} +y^{2}+4 x-1=0$

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2 years ago