Answer:
The equation of the circle in standard form if the points (3,5) and (-5,-1) are endpoints of the diameter is
[tex]x^{2}+y^{2} +2x-4y-20=0[/tex]
Step-by-step explanation:
Let AB be the Diameter and the end points be
point A( x₁ , y₁) ≡ ( 3 , 5)
point B( x₂ , y₂) ≡ (-5 , -1)
A circle is given as AB is a Diameter.
To Find:
Equation of the circle in standard form = ?
Solution:
Diameter form of the equation of the circle is
(x - x₁)(x - x₂) + (y - y₁)(y - y₂ ) = 0
Where ( x₁ , y₁) and ( x₂ , y₂) are the endpoints of the diameter.
So on substituting the value we get the required equation of circle
(x - 3)(x - (-5)) + (y - 5)(y - (-1)) = 0
∴ (x - 3)(x + 5) + (y - 5)(y + 1) = 0
Applying Distributive property we get
[tex]x(x+5)-3(x+5)+y(y+1)-5(y+1)=0\\\\x^{2}+5x-3x-15+y^{2}+y-5y-5=0\\x^{2}+y^{2} +2x-4y-20=0\\\therefore x^{2}+y^{2} +2x-4y-20=0\ \textrm{which is the required equation of the circle}[/tex]
The equation of the circle in standard form if the points (3,5) and (-5,-1) are endpoints of the diameter is
[tex]x^{2}+y^{2} +2x-4y-20=0[/tex]
