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WARRIOR [948]
3 years ago
14

The edges of the diameter of a circle are at (-1,2) and (3,-1). Write the equation of the circle in general form.

Mathematics
1 answer:
Y_Kistochka [10]3 years ago
3 0

Answer:

(x-1)²+ (y-0.5)²=6.25

Step-by-step explanation:

<u>The standard form of equation of a circle is;</u>

(x-a)²+(y-b)²=r² where (a,b) are the center of the circle and r is the radius

<u>Finding the mid-point of the given points</u>

(-1,2) and (3,-1)⇒midpoint will be 1/2(x₁+x₂) , 1/2(y₁+y₂)

midpoint= {1/2(-1+3), 1/2(2+-1)}

midpoint=(1,0.5)

<u>Finding the radius r; the distance from the center to either of the given two points</u>

Apply the distance formula d=√ (x₂-x₁)² +(y₂-y₁)²

Taking (x₁,y₁) as (1,0.5) and (x₂,y₂) as (-1,2) then

d=√ (-1-1)² +(2-0.5)²

d= √ (-2)²+(1.5)²

d=√4+2.25⇒√6.25⇒2.5

r=2.5

<u>Equation of the circle</u>

(x-1)² + (y-0.5)²=2.5²

(x-1)²+ (y-0.5)²=6.25

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Step-by-step explanation:

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m(∠AED) = 180° - 110°

m(∠AED) = 70°

Since, ∠AED ≅ ∠EAD

Therefore, m∠AED = m∠EAD = 70°

b). By triangle sum theorem in ΔABD,

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110° + 40° + m∠DAB = 180°

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m∠DAB = 30°

m∠BAE = m∠EAD + m∠BAD

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m∠EAC + m∠AEC + m∠ACE = 180°

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mr_godi [17]

See the attached figure.

m ∠KAJ = 170° &  m ∠LAM = 80°

We should know that :

m ∠KAJ + m ∠LAM + m ∠KAL + m ∠MAJ = 360°

∴ m ∠KAL + m ∠MAJ = 360° - (m ∠KAJ + m ∠LAM)

∴ m ∠KAL + m ∠MAJ = 360° - (170°+80°) = 360° - 250° = 110°

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∴ m ∠MAJ + m ∠MAJ = 110°

∴ 2 * m ∠MAJ = 110°

∴ m ∠MAJ = 110° ÷ 2 = 55°


<u>So, the answer is :  m ∠MAJ = 55°</u>

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