Question

Question #754ab

Answer 1

`(x - 3/2)^2 + (y+1/2)^2 = 5/2`

Explanation:

The equation of a circle in standard form:
`(x - h)^2 + (y - k)^2 = r^2`, where `"center" = (h,k), "radius" = r`

Find the radius:

The diameter of the circle is the distance between the two points:
`d = sqrt((-1 - 0)^2 + (3-0)^2) = sqrt(1 + 9) = sqrt(10)`

`r = d/2 = sqrt(10)/2;" " r^2 = (sqrt(10)/2)^2 = 10/4 = 5/2`

Find the center:
The center is the midpoint of the diameter `((x_1 +x_2)/2, (y_1 + y_2)/2) = (3/2, -1/2)`:

The circle is required to be tangent to the line `3x + y = 9`. The radius and the tangent line are perpendicular.

The line in `y = mx + b` form:

`y = -3x + 9` with `m = -3`

`m_("perpendicular") = 1/3`

The perpendicular line that goes through the diameter is:
`y - 0 = 1/3(x - 3) => y = 1/3x - 1`

The point `(0, -1)` is on this line!

Therefore, the equation of the circle is:
`(x - 3/2)^2 + (y+1/2)^2 = 5/2`

Answer 2

`(x-3/2)^2+(y+1/2)^2=5/2,`

or,

`4x^2+4y^2-12x+4y=0.`

A little correction : the eqn. is `x^2+y^2-3x+y=0.`

Explanation:

We will deal with the Problem using the Geometrical Concepts.

The points `A(3,0) and B(0,-1)` are on the Circle. So, the

sgmnt. `AB` is a chord of the circle. From Geometry, we know

that the Centre `C` lies in the `bot-`bisector of a chord....(1).

Let, line `l_1` be the `bot-`bisct. of the chord `AB`,

We also know that, the `bot` to a Tangent at the Point of Contact

passes thro. the Centre of the circle......(2).

Let, line `l_2` be the `bot` to the tgt. line `t : 3x+y-9=0` at the

pt. of cot. `T(3,0).`

From (1) and (2), we find, `l_1nnl_2={C}........(star).`

Eqn. of `l_1`:-

Slope of `AB` is, `(0-(-1))/(3-0)=1/3 :." slpoe of "l_1" is "-3.`

Mid-pt. `M` of `AB`, i.e., `M((0+3)/2,(-1+0)/2)=M(3/2,-1/2) in l_1.`

`:. l_1 : y-(-1/2)=-3(x-3/2),`

`or, l_1 : y+1/2=-3(x-3/2)....(1').`

On the similar lines, we get, `l_2 : y-0=1/3(x-3)......(2').`

Solving `(1') and (2')` and, from `(star)`, we get `C(3/2,-1/2).`

If `r` is the Radius of the circle, then,

`r^2=BC^2=(3/2)^2+(1/2)^2=5/2.`

Thus, the centre of the circle is `C(3/2,-1/2), and, r^2=5/2.`

Hence, the eqn. of the reqd. circle is,

`(x-3/2)^2+(y+1/2)^2=5/2,` or,

`4x^2+4y^2-12x+4y=0.`

A little correction : the eqn. is `x^2+y^2-3x+y=0.`

Enjoy Maths.!