Please answer these math sums (coordinate geometry- circle)?

Question

1) Find the equation of a circle whose center is on 2x+5y=1 and passing through (2,3) and (1,5)

2) Find the equation of a circle which is concentric with the circle x^2 + y^2 + 4x - 6y + 11 = 0 and passes through (2,4)

Thank you..

Answer 1

1)

To find the center of the circle, we can use the fact that the distance between the line and the two points must be the same.

Re-arranging 2x + 5y = 1 to slope-intercept form gives y = -2x/5 + 1/5.

By the Distance Formula, the distance between (x, -2x/5 + 1/5) and (2, 3) is:

D = √[(x - 2)^2 + (-2x/5 + 1/5 - 3)^2].

The distance between (x, -2x/5 + 1/5) and (1, 5) is:

D = √[(x - 1)^2 + (-2x/5 + 1/5 - 5)^2].

Equating the distances:

√[(x - 2)^2 + (-2x/5 + 1/5 - 3)^2] = √[(x - 1)^2 + (-2x/5 + 1/5 - 5)^2]
==> (x - 2)^2 + (-2x/5 + 1/5 - 3)^2 = (x - 1)^2 + (-2x/5 + 1/5 - 5)^2
==> 18x + 61 = 0
==> x = -61/18.

This gives the y-coordinate of the center to be -2(-61/18)/5 + 1/5 = 14/9.

Since the square of the distance between (-61/18, 14/9) and (2, 3) is (2 + 61/18)^2 + (3 - 14/9)^2 = 10085/324, the required equation is:

(x + 61/18)^2 + (y - 14/9)^2 = 10085/324.

2)

The given circle in standard form is:

(x^2 + 4x) + (y^2 - 6y) = -11
==> (x^2 + 4x + 4) + (y^2 - 6y + 9) = 2
==> (x + 2)^2 + (y - 3)^2 = 2.

Thus, the circle we need passes through (2, 4) and has a center of (-2, 3).

The circle takes the form:

(x + 2)^2 + (y - 3)^2 = r^2.

Since (2, 4) passes through it, we have:

(2 + 2)^2 + (4 - 3)^2 = r^2
==> r^2 = 17.

Therefore, the required circle is (x + 2)^2 + (y - 3)^2 = 17.

I hope this helps!

Answer 2

Several ways to solve this problem

1) the equation of circle is

( x-a)^2 + (y-b)^2 =r^2 (eq1 )

Plug in (2,3) into eq1 you have eq2

Plug in (1,5) into eq1 you have eq3

Besides, 2a + 5b =1 (eq4)

Solve (eq2) eq3 , eq3 simultaneously for a,b ,r

I got a = - 61/ 18 , b =14/9 and r = 5.579

Answer 3

1. 6

2. 82