EQUATION OF CIRCLE WHEN IT PASSES THROUGH INTERSECTION OF LINES

Example 1 :

Find the equation of the circle with center (2, 3) and passing through the intersection of the lines 3x − 2y −1 = 0 and 4x + y − 27 = 0

Solution :

By finding the point of intersection of the lines, we get a point lies on the circle.

3x − 2y − 1 = 0  ----------(1)

4x + y − 27 = 0 ----------(2)

                  3x − 2y − 1 = 0

(2) x 2 ==> 8x + 2y − 54 = 0

                --------------------

               11x - 55  =  0  ==> x  =  5

Apply x = 5 in (1) to get the value of y.

3(5) - 2y - 1  =  0

15 - 2y - 1  =  0

-2y  =  -14

y  =  7

So, the point lies on the circle is (5, 7).

By finding the distance between center and a point lies on the circle, we get the radius.

r = √(2 - 5)² + (3 - 7)²

r = √(-3)² + (-4)²   =  √25

r = 5

Equation of the circle :

(x - h)² + (y - k)²  =  r² 

(h, k)  ==>  (2, 3)

(x - 2)² + (y - 3)²  =  5² 

x² - 4x + 4 + y² - 6y + 9 - 25  =  0

x² + y² - 4x - 6y - 12  =  0

Equation of the Circle with Endpoints of Diameter are Given

Example 2 :

Obtain the equation of the circle for which (3, 4) and (2,-7) are the ends of a diameter.

Solution :

Equation of circle with end points of diameter :

(x − x₁)(x − x₂) + ( y − y₁)( y − y₂)  = 0

(x₁, y₁)  ==>  (3, 4) and (x₂, y₂)  ==>  (2, -7)

(x − 3)(x − 2) + ( y − 4)( y + 7)  = 0

x² - 5x + 6 + y² + 3y - 28  =  0

x² + y² - 5x + 3y + 6 -28  =  0

x² + y² - 5x + 3y - 22  =  0

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