Question

find the equation of a circle passing through origin and the centre is (7,-2)​

Answer 1

Answer :

x² + y² - 14x + 4y = 0

Solution :

Here ,

We need to find the equation of the circle which passes through origin and has the centre at (7 , -2) .

Firstly ,

Let's find the radius of the circle .

Since (7 , -2) is the centre and (0 , 0) is a point on circumference of the circle ( °•° circle passes through the origin) , thus using the distance formula , the radius r of the circle will be ;

=> r = √[ (7 - 0)² + (-2 - 0)²]

=> r = √[ 7² + (-2)²]

=> r = √[ 49 + 4 ]

=> r = √53

Also ,

We know that , the standard form of a circle with center at (h , k) and radius r is given as ;

(x - h)² + (y - k)² = r²

Thus ,

The equation of the circle under consideration whose center is at (7 , -2) and radius √53 will be given as ;

=> (x - 7)² + (y - (-2))² = (√53)²

=> (x - 7)² + (y + 2)² = 53

=> x² - 14x + 49 + y² + 4y + 4 = 53

=> x² + y² - 14x + 4y + 53 = 53

=> x² + y² - 14x + 4y + 53 - 53 = 0

=> x² + y² - 14x + 4y = 0

Hence ,

The required equation is ;

x² + y² - 14x + 4y = 0