Find the equation of the circle which passes through the point (1, 4) and whose equation of two diameters are x – y = 1 and 2x + 3y = 7
Answers
Answer:
EXPLANATION.
Equation of circle which passes through the point (1,4).
Whose equation of two diameters are,
⇒ x - y = 1 and 2x + 3y = 7.
As we know that,
First we find the point of intersection of two diameter.
⇒ x - y = 1. - - - - - (1).
⇒ x = 1 + y. - - - - - (1).
⇒ 2x + 3y = 7. - - - - - (2).
Put the values of equation (1) in equation (2), we get.
⇒ 2(1 + y) + 3y = 7.
⇒ 2 + 2y + 3y = 7.
⇒ 5y + 2 = 7.
⇒ 5y = 7 - 2.
⇒ 5y = 5.
⇒ y = 1.
Put the value of y = 1 in equation (1), we get.
⇒ x = 1 + y.
⇒ x = 1 + 1.
⇒ x = 2.
Their Co-ordinates = (2,1).
Centre of circle : (2,1).
Radius is Distance between (1, 4) and (2,1).
As we know that,
Distance formula :
⇒ d = √(x₂ - x₁)² + (y₂ - y₁)².
Put the values in the equation, we get.
⇒ d = √(2 - 1)² + (1 - 4)².
⇒ d = √(1)² + (-3)².
⇒ d = √1 + 9.
⇒ d = √10.
As we know that,
Equation of circle.
⇒ (x - h)² + (y - k)² = r².
Put the values in the equation, we get.
⇒ (x - 2)² + (y - 1)² = (√10)².
⇒ x² + 4 - 4x + y² + 1 - 2y = 10.
⇒ x² + y² - 4x - 2y + 5 = 10.
⇒ x² + y² - 4x - 2y + 5 - 10 = 0.
⇒ x² + y² - 4x - 2y - 5 = 0.