Question

Find the equation of circum-cirle of the triangle formed by the straight lines given the following



pls give me correct explanation​

Answer 1

Step-by-step explanation:

subtract eq: 1 from the 2

and then reminder eq: from eq:3

you get the answer.

Answer 2

Answer:

7(x² + y²) + 40x – 37y + 35 = 0

Step-by-step explanation:

Let us assume that the given lines form a triangle ABC. So, Let,

AB : 5x – 3y + 4 = 0  

AC : 2x + 3y – 5 = 0  

BC : x + y = 0

Solving these 3 equations, we get,

$A = \bigg(\dfrac{1}{7} , \dfrac{11}{7} \bigg) \ ; \ B = \bigg(-\dfrac{1}{2} , \dfrac{1}{2} \bigg) \ ; \ C = (-5, 5)$

A, B, C are the vertices of the triangle ABC and these are its coordinates.

We know that, equation of the circle is

x² + y² + 2gx + 2fy + c = 0

Points A, B, C are on circumference of circle

So, substituting this in the circle equation, we get

Substituting A,

$\dfrac{1}{49}+\dfrac{121}{49}+\dfrac{2g}{7} + \dfrac{22f}{7} + c = 0$

$: \longmapsto 14g + 44f + 49c = -122 \ \ \ ---[1]$

Substituting B,

$\dfrac{1}{4} +\dfrac{1}{4} -g+f+c = 0$

$: \longmapsto 2g - 2f - 2c = 1 \ \ \ ---[2]$

Substituting C,

$25 + 25 -10g+10f + c = 0 \ \ \ ---[3]$

Solving, [1] , [2] and [3], we get

$g = \dfrac{20}{7} \ ; \ f = -\dfrac{37}{14} \ ; \ c = 5$

So, substituting these values in the circle equation, we get,

$x^2 + y^2 + \dfrac{40}{7} x -\dfrac{37}{7} y + 5 = 0$

$: \longmapsto \boxed{\boxed{7(x^2+y^2) +40x -37y+35 = 0}}$

is the equation of given circle.

Hope it helps!!