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
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Answer 1

Answer:

Step-by-step explanation:

The equation of two diameters of the circle is

$\sf{x-y=1}\\\sf{2x+3y=7}$

So, the intersection of the diameters will be the center of the circle

The center of the circle ≡ (2 , 1)

Now, the equation of the circle with center (2,1) is

$\rm{\left(x-2\right)^{2}+\left(y-1\right)^{2}={r}^{2}}$

It is given that the circle is passing through (1,4), so,

$\rm{\left(1-2\right)^{2}+\left(4-1\right)^{2}={r}^{2}}$

$\rm{\implies\,\left(-1\right)^{2}+\left(3\right)^{2}={r}^{2}}$

$\rm{\implies\,1+9={r}^{2}}$

$\rm{\implies\,{r}^{2}=10}$

Now, the equation of the circle becomes,

$\rm{\left(x-2\right)^{2}+\left(y-1\right)^{2}=10}$

$\rm{\implies\,{x}^{2}-4x+4+{y}^{2}-2y+1=10}$

$\rm{\implies\,{x}^{2}+{y}^{2}-4x-2y+5=10}$

$\rm{\implies\,{x}^{2}+{y}^{2}-4x-2y-5=0}$

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