85.0k views
1 vote
find the equation of the circle with centre (2,3) and passing through the intersection of the lines 3x-2y-1=0 and 4x+y-27=0​

1 Answer

2 votes

Answer:

3x-2y-1=0

or, 3x-2y=1

or, x=(1+2y)/3...(i)

4x+y=27... ...(ii)

solving (i) and (ii)

(4+8y)+3y=27x3

4+11y= 81

11y=77

y=7

now,

x=(1+2x7)/3

x=5

hence, the passing point is (5,7)

now, for the radius of circle

radius of circle= distance betn (5,7) and (2,3)

r= √(5-2)²+(7-3)²

r=5units

finally,

the eqn of circle

(x-h)²+(y-k)²=r² (where (h,k)=(2,3) )

(x-2)²+ (y-3)²=5²

x²-4x+4+y²-6y+9=25

x²+y²-4x-6y=12....this is the required answer.

User TantanQi
by
8.3k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.