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satela [25.4K]
3 years ago
9

The center of a circle is located at (3, 8), and the circle has a radius that is 5 units long. What is the general form of the e

quation for the circle?
Mathematics
1 answer:
n200080 [17]3 years ago
4 0
Center of the circle: C=(3,8)=(h,k)→h=3, k=8
Radius of the circle: r=5
Standard form of the equation for the circle:
(x-h)^2+(y-k)^2=r^2
(x-3)^2+(y-8)^2=5^2
(x-3)^2+(y-8)^2=25

General for of the equation for the circle:
(x)^2-2(x)(3)+(3)^2+(y)^2-2(y)(8)+(8)^2=25
x^2-6x+9+y^2-16y+64=25
x^2+y^2-6x-16y+73=25
x^2+y^2-6x-16y+73-25=25-25
x^2+y^2-6x-16y+48=0

Answer: The general form of the equation for the circle is:
x^2+y^2-6x-16y+48=0
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The sum of the angles of a triangle is 180°. If one angle of a triangle measures x and the second angle measures
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the third angle= 180-x-(4x+15)

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3 years ago
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5 0
3 years ago
O is the centre of the circle, EF is a tangent, angle BCE = 28°, angle ACD = 31°
andriy [413]

Answer

a. 28˚

b. 76˚

c. 104˚

d. 56˚

Step-by-step explanation

Given,

∠BCE=28°  ∠ACD=31°  &  line AB=AC .

According To the Question,

  • a. the angle between a chord and a tangent through one of the end points of the chord is equal to the angle in the alternate segment.(Alternate Segment Theorem) Thus, ∠BAC=28°

  • b. We Know The Sum Of All Angles in a triangle is 180˚, 180°-∠CAB(28°)=152° and ΔABC is an isosceles triangle, So 152°/2=76˚

        thus , ∠ABC=76° .

  • c. We know the Sum of all angles in a triangle is 180° and opposite angles in a cyclic quadrilateral(ABCD) add up to 180˚,

Thus, ∠ACD + ∠ACB = 31° + 76° ⇔ 107°

Now, ∠DCB + ∠DAB = 180°(Cyclic Quadrilateral opposite angle)

∠DAB = 180° - 107° ⇔ 73°

& We Know, ∠DAC+∠CAB=∠DAB ⇔ ∠DAC = 73° - 28° ⇔ 45°

Now, In Triangle ADC Sum of angles in a triangle is 180°

∠ADC = 180° - (31° + 45°)  ⇔  104˚

   

  • d. ∠COB = 28°×2 ⇔ 56˚ , because With the Same Arc(CB) The Angle at circumference are half of the angle at the centre  

For Diagram, Please Find in Attachment  

4 0
3 years ago
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