What is the equation of a circle with center (-3, -1) that contains the point (1, 2)?
1 answer:
So, we know the center is at -3,-1, ok
hmmm what's the radius anyway? well, we know that there's a point at 1,2 that is on the circle's path...hmmmm what's the distance from the center to that point? well, is the radius, let's check then.
![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -3}}\quad ,&{{ -1}})\quad % (c,d) &({{ 1}}\quad ,&{{ 2}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ r=\sqrt{[1-(-3)]^2+[2-(-1)]^2}\implies r=\sqrt{(1+3)^2+(2+1)^2} \\\\\\ r=\sqrt{16+9}\implies r=\sqrt{25}\implies r=5](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%28%7B%7B%20-3%7D%7D%5Cquad%20%2C%26%7B%7B%20-1%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0A%26%28%7B%7B%201%7D%7D%5Cquad%20%2C%26%7B%7B%202%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ar%3D%5Csqrt%7B%5B1-%28-3%29%5D%5E2%2B%5B2-%28-1%29%5D%5E2%7D%5Cimplies%20r%3D%5Csqrt%7B%281%2B3%29%5E2%2B%282%2B1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ar%3D%5Csqrt%7B16%2B9%7D%5Cimplies%20r%3D%5Csqrt%7B25%7D%5Cimplies%20r%3D5)
so, what's the equation of a circle with center at -3, -1 and a radius of 5?
![\bf \textit{equation of a circle}\\\\ (x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2 \qquad \begin{array}{lllll} center\ (&{{ h}},&{{\quad k}})\qquad radius=&{{ r}}\\ &-3&-1&5 \end{array} \\\\\\\ [x-(-3)]^2+[y-(-1)]^2=5^2\implies (x+3)^2+(y+1)^2=25](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%7B%7B%20h%7D%7D%29%5E2%2B%28y-%7B%7B%20k%7D%7D%29%5E2%3D%7B%7B%20r%7D%7D%5E2%0A%5Cqquad%20%0A%5Cbegin%7Barray%7D%7Blllll%7D%0Acenter%5C%20%28%26%7B%7B%20h%7D%7D%2C%26%7B%7B%5Cquad%20%20k%7D%7D%29%5Cqquad%20%0Aradius%3D%26%7B%7B%20r%7D%7D%5C%5C%0A%26-3%26-1%265%0A%5Cend%7Barray%7D%20%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bx-%28-3%29%5D%5E2%2B%5By-%28-1%29%5D%5E2%3D5%5E2%5Cimplies%20%28x%2B3%29%5E2%2B%28y%2B1%29%5E2%3D25)
hmmm what's the radius anyway? well, we know that there's a point at 1,2 that is on the circle's path...hmmmm what's the distance from the center to that point? well, is the radius, let's check then.
so, what's the equation of a circle with center at -3, -1 and a radius of 5?
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The area of a right angled triangle with sides of length 9cm, 12cm and 15cm in square centimeters is 54 sq cm.
The formula to calculate the area of a right triangle is given by:
Area of Right Triangle, A = (½) × b × h square units
Where, “b” is the base (adjacent side) and “h” is the height (perpendicular side). Hence, the area of the right triangle is the product of base and height and then divide the product by 2.
We know that the hypotenuse is the longest side. So, the area of a right angled triangle will be half of the product of the remaining two sides.
Given sides of the triangle:
a=9cm
b=12cm
c=15cm
From this we know that the hypotenuse is c. Are of the triangle will be obtained by the other two sides.
∴Area = x 9 x 12
= 54
Answer:
Step-by-step explanation:
The complete question is provided in the attachment.
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