Question

Find the equation of the circle passing through the origin and has centre at (-4,-3)​

Answer 1

Step-by-step explanation:

1. if line pass origin so coordinate is( 0,0) and second coordinate (-4,-3) and we know the formula
2. $y - y1 = y2 - y1 \div x2 - x1(x - x1$
3. so the value is x1 is 0 and y1 is also 0 and x2 is -4 and y2 is -3
4. so put the value
5. y-0=-3-0÷-4 -0=x-0
6. y=3÷-4=x
7. -4y=3x
8. 3x+4y=0

Answer 2

Answer:

x² + y² + 8x + 6y =0

Step-by-step explanation:

Concept= Equation of Circle

Given= Circle passing through origin and the point of center

To Find= The equation of the Circle

Explanation=

We have been given the question as to find the equation of the circle passing through the origin and has center at (-4,-3)​.

So we have the point of center as (-4,-3) and the center as (0,0).

Radius of circle is the distance between the center and the origin.

r= √{(-4-0)² + (-3-0)²}

=> √{(-4)² + (-3)²}

=> √{ 16 + 9}

=> √25 = 5 units

therefore the radius of circle is 5 units.

And we know that the equation of circle passing through origin (0,0) and having center as (p,q) with radius r is

(x-p)² + (y-q)² =r²

So our (p,q) is (-4,-3) and r=5

The equation of circle is

{x-(-4)}² + {y-(-3)}² = 5²

{x+4}² + {y+3}² = 25

x² +16 + 8x + y² + 9 + 6y = 25

x² + y² + 8x + 6y +25=25

x² + y² + 8x + 6y =0.

Therefore the equation of circle passing through origin and the center (-4,-3) is x² + y² + 8x + 6y =0.

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