Question

find the equation of the circle of which the line joining the origin and the point (2,-4)is a diameter

Answer 1

Answer:

x^2 + y^2 - 2x + 4y = 0

Step-by-step explanation:

Given end points of the diameter: (x1, y1) = (0, 0) and (x2, y2) = (2, -4)

The mid-point of the line passing through the given points is the centre of the circle, (h,k)

(h,k) = ((x1+x2)/2, (y1+y2)/2)

(h,k) = ((0+2)/2, (0+(-4))/2) = (1,-2)

The radius of the circle is given by the distance between centre of the circle and any one of the end points of the diameter.

The formula for the distance between two points:  

√((x2 - x1)^2+ (y2 - y1)^2 ) = r

Hence, the distance between one of the given end points (0, 0) and the centre of the circle (1, -2) is:

r = √((1-0)^2 + (-2-0)^2 ) = √5

∴ radius of the given circle, r = √5

The equation of the circle with centre (h,k) and radius, r is:

  (x-h)^2 + (y-k)^2 = r^2

  (x-1)^2 + (y-(-2))^2 = (√5)^2

  (x^2 - 2x + 1) + (y^2 + 4y + 4) = 5  [∵ (a – b)^2 = a^2–2ab+b^2]

  (x^2 + y^2 – 2x + 4y + 5) – 5 = 0  [∵ taking the 5 on RHS to the LHS]

∴ The equation of the given circle is  x^2 + y^2 - 2x + 4y = 0  

Answer 2

Solution:

Concept:

The equation of circle having endoints of diamter $(x_1,y_1)\:and\:(x_2,y_2)$is

$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

As per given data,

the endpoints of diameter are (0,0) and (2,-4)

The equation of the required circle is

$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

(x-0)(x-2)+(y-0)(y+4)=0

x(x-2)+y(y+4)= 0

x² -2x + y² + 4y = 0

x² + y² -2x + 4y = 0