Question

find the equation of the circle which passes through the points(4 , 3) and (-2 , 5) and having its centre in the line 2x-3y =4... plz ans this math.​

Answer 1

Answer:

$x^{2}+y^{2}+2x+4y-5=0$

Step-by-step explanation:

We know that-

The equation of a circle is-

$x^{2}+y^{2}+2gx+2fy+c=0$   ---------------------- A

Since the circle passing through the points (4,3) and (-2,5) then these points lie on the circle. So,

For point (4,3)

$4^{2}+3^{2}+2g.4+2f.3+c=0$

16 + 9 + 8g + 6f + c = 0    

25 + 8g + 6f + c = 0

8g + 6f + c = -25           ------------------------- 1

For point (-2,5) -

$(-2)^{2}+5^{2}+2g(-2)+2f.5+c=0$

4 + 25 - 4g + 10f + c = 0

29 -4g + 10f + c = 0

-4g + 10f + c = -29              -------------------- 2

Since the centre (-g,-f) lies on the line 2x - 3y = 4

Then 2(-g) -3(-f) = 4

-2g + 3f = 4     ---------------- 3

Subtracting equation 1 and 2, we have-

8g + 6f + c = -25

-4g + 10f + c = -29  

Subtracting-

12g - 4f = 4            --------------------------- 4

Solving equation 3 and 4

-2g + 3f = 4     ---------------- 3

12g - 4f = 4            --------------------------- 4

Multiply in equation 3 by 6 and adding equation 3 and 4

-12g + 18f = 24

12g - 4f = 4    

Adding-

14f = 28

f = $\dfrac{28}{14}$

f = 2

Put f = 2 in equation 3-

-2g + 3f = 4

-2g + 3(2) = 4

-2g + 6 = 4

-2g = -2

2g = 2

g = 1

Put g = 1 and f = 2 in equation 1

8g + 6f + c = -25

8(1) + 6(2) + c = -25

8 + 12 + c = -25

20 + c = -25

c = -5

Put the values of g, f and c in equation A

$x^{2}+y^{2}+2gx+2fy+c=0$

$x^{2}+y^{2}+2x(1)+2y(2)+(-5) = 0$

$x^{2}+y^{2}+2x+4y-5=0$

This is the required equation of circle.