How do you write an equation of a circle with Center: (6, 7) Point on Circle: (13 4)?

2 Answers
Jan 17, 2017

Answer:

#x^2+y^2-12x-14y+27=0#

Explanation:

Let say #x and y# is another point on the circle.

#(x-6)^2+(y-7)^2=(13-6)^2+(4-7)^2#
#x^2-12x+36+y^2-14y+49=(7)^2+(-3)^2#
#x^2-12x+36+y^2-14y+49=49+9#
#x^2+y^2-12x-14y+36+49-49-9=0#
#x^2+y^2-12x-14y+27=0#

Jan 17, 2017

Answer:

#x^2+y^2-12x-14y+27=0.#

Explanation:

The Centre #C# of the Circle is #C(6,7)#

The Point #P(13,4)# is on the Circle. Therefore, the Radius #r# of the

Circle is distance #CP#. Using, the Distance Formula,

#r^2=CP^2=(13-6)^2+(4-7)^2=49+9=58.#

We know that the eqn. of a Circle having Centre at #(h,k) and Radius

#r" "is : (x-h)^2+(y-k)^2=r^2#

Therefore, the eqn. of the circle is

#(x-6)^2+(y-7)^2=58, i.e.,#

#x^2+y^2-12x-14y+27=0#

Spread the Joy of Maths.!