Question

Find the equation of the circle which passes through the point (1, 4) and whose equation of two diameters are x – y = 1 and 2x + 3y = 7 thankyou sis or bro
what is your name​

Answer 1

Answer:

The diameter is a chord of the circle, which goes through the center of the circle. I am assuming that (x- y) =1 is the equation of the diameter, when extended as a line. The intersection of (x-y)=1 and (2x+3y) =7 is the center of the circle. Solving the 2 eqns, 5y = 5. y = 1, x = 2. The equation of a circle with center (2, 1) is (x-2)^2 + (y-1)^2 = radius^2. Since (1,4) is on its circumference, radius^2 = 1 + 9 = 10.

The equation of the desired circle is (x-2)^2 +(y-1)^2 = 10.

Step-by-step explanation:

my name is LAKSHAY

pls mark branliest pls I am waiting

Answer 2

Answer:

(x − 2) 2 + (y − 1) 2 = 10

Step-by-step explanation:

How do you find the equation of a circle which passes through (1,4) and has diameters on the lines given by x−y=1 and 2x+3y=7?

Since those lines are not parallel, and include diameter of the circle, they must intersect at the center of the circle. We can find that intersection by substituting one equation into the other:

x=y+1

2 (y + 1) + 3y = 7⟹5y=5⟹y=1

x = 1 + 1 = 2

Thus, the center is at (2,1).

Now, we need the radius (or radius squared, really.) That’s the distance between the center (2,1) and given point (1,4):

(2 − 1) 2 + (1 − 4) 2 = r2⟹10=r2

Thus, our circle is given by:

(x − 2) 2 + (y − 1) 2 = 10