Question

please help i am little confused
both questions 3 and 4

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thanks and all the best .

Answer 1

hii mate

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here is ur answer ^↓^

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Q³
let us replace values of x and y in all the equations.
x=4
y=5

a) x²+y²+4x+4y-5=0
→4²+5²+4(4)+4(5)-5=0
→16+25+16+20-5=0
→32+40=0
→72≠0
✖✖rejected✖✖
••••••••••••••••••••

b)x²+y²-4x-4y-5=0
→4²+5²-4(4)-4(5)-5=0
→16+25-16-20-5=0
→16-16+25-25=0
→0+0=0
→0=0
✨Selected✨✔✔✔✔
′′′′′′′′′′′′′′′′′′′′′′′′′′′′

the equation for 3rd is x²+y²-4x-4y-5=0

______________________

Q⁴
x²+y²-12x+4y+6=0

let us consider

a) x+y=0
so,x=(-y)

→(-y)²+y²-12(-y)+4y+6=0
→y²+y²+12y+4y+6=0
→2y²+16y+6=0
→2(y²+8y+3)=0
→y²+8y+3=0/2{=0}
since the equation has no real roots {check by split method}

✖✖rejected✖✖
•••••••••••••••••••

b) x+3y=0
x=(-3y)

→(-3y)²+y²-12(-3y)+4y+6=0
→9y²+y²+36y+4y+6=0
→10y²+40y+6=0
→2(5y²+20y+3)=0
→5y²+20y+3=0

no real roots exists{check by split method}

✖✖rejected✖✖
•••••••••••••••••••

c)x=y

→y²+y²-12(y)+4y+6=0
→2y²-12y+4y+6=0
→2(y²-2y+3)=0
→y²-2y+3=0

using split method
→-y²-3y+1y+3=0.......consider y² as -y² as square of any no. is +ve.
→-y(1y+3)+1(y+3)=0
→(-y+1)(y+3)=0

values of y are 1,-3

since, real roots exists

✨selected✨✔✔✔
′′′′′′′′′′′′′′′′′′′′′′′′′′′

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⭐hope it helps uh⭐

✨_____☺_____✨

please mark it as brainliest✌✌✌✌✌

Answer 2

(3)

We know that Equation of circle is (x - a)² + (y - b)² = r².

Here, (a,b) is the centre of the circle. radius is the distance between the centre of circle and any point on the centre of circle.

Given centre of circle is (2,2) and equation of circle is passing through (4,5)

Here, (x₁,y₁) = (2,2) and (x₂,y₂) = (4,5).

Now,

We know that distance between the two points is calculated as:

⇒ √(x₂ - x₁)² + (y₂ - y₁)²

⇒ √(4 - 2)² + (5 - 2)²

⇒ √(2)² + (3)²

⇒ √4 + 9

⇒ √13.

Substitute the value in (1), we get

⇒ (x - 2)² + (y - 2)² = (√13)²

⇒ x² + 4 - 4x + y² + 4 - 4y = 13

⇒ x² + y² - 4x - 4y + 8 = 13

⇒ x² + y² - 4x - 4y = 5

⇒ x² + y² - 4x - 4y - 5 = 0.

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(14)

Given Equation is x^2 + y^2 - 12x + 4y + 6 = 0.

Option Verification:

(a)

⇒ x + y = 0

⇒ x = -y.

Substitute in (1),we get

⇒ (y)^2 + (y)^2 + 12y + 4y + 6 = 0

⇒ 2y^2 + 16y + 6 = 0

⇒ y^2 + 8y + 3 = 0

⇒ y = √13 - 4, y = -4 - √13

(b)

⇒ x + 3y = 0

⇒ x = -3y.

Substitute in (1),we get

⇒ (-3y)^2 + y^2 - 12(-3y) + 4y + 6 = 0

⇒ 9y^2 + y^2 + 36y + 4y + 6 = 0

⇒ 10y^2 + 40y + 6 = 0

⇒ 5y^2 + 20y + 3 = 0

⇒ y =√85 - 10/5, y = -10 + √85/3

(iii)

x = y

⇒ (y)^2 + (y)^2 - 12(y) + 4y + 6 = 0

⇒ 2y^2 - 12y + 4y + 6 = 0

⇒ 2y^2 - 8y + 6 = 0

⇒ y^2 - 4y + 3 = 0

⇒ y^2 - y - 3y + 3 = 0

⇒ y(y - 1) - 3(y - 1) = 0

⇒ (y - 3)(y - 1) = 0

⇒ y = 3,1.

(iv)

⇒ 3x + 2y = 0

⇒ 3x = -2y

⇒ x = -2y/3

Substitute in (1), we get

⇒ (-2y/3)^2 + y^2 - 12(-2y/3) + 4y + 6 = 0

⇒ (13y^2/9) + 12y + 6 = 0

⇒ y = 3(√246 - 18)/13, -3(18 + √246)/13.

Therefore, the answer is x = y.

Hope this helps!