Question #754ab

2 Answers
Apr 7, 2017

#(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!

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

Apr 7, 2017

#(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.!