Mathematics: 20, 21, 28 acute obtuse right

20, 21, 28 acute obtuse right - id20212812529929, xmhllo

Ответ:

corianrigby97

26.10.2021

Mathematics

$(x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}$

Step-by-step explanation:

Given:

Center of circle is at (5, -4).

A point on the circle is (x_1,y_1)=(-3, 2)

Equation of a circle with center (h,k) and radius 'r' is given as:

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

Here, (h,k)=(5,-4)

Radius of a circle is equal to the distance of point on the circle from the center of the circle and is given using the distance formula for square of the distance as:

r^2=(h-x_1)^2+(k-y_1)^2

Using distance formula for the points (5, -4) and (-3, 2), we get

$r^2=(5-(-3))^2+(-4-2)^2\\r^2=(5+3)^2+(-6)^2\\r^2=8^2+6^2\\r^2=64+36=100$

Therefore, the equation of the circle is:

$(x-5)^2+(y-(-4))^2=100\\(x-5)^2+(y+4)^2=100$

Now, rewriting it in the form asked in the question, we get

$(x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}$

0,0(0 оценок)

20, 21, 28 acute obtuse right

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question

20, 21, 28 acute obtuse right​

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question

20, 21, 28 acute obtuse right