Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the center lies on the straight line y – 4x + 3 = 0.

Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line y – 4x + 3 = 0.

Answers (1)

Let the centre of the circle be C(h,k)

 Given that the centre lies on the line y-4x+3=0

y-4x+3=0

k-4h+3=0

k=4h-3

AC2=BC2

(h-2)2+(4h-3-3)2=(h-4)2+(4h-3-5)2

h2-4h+4+16h2-48h+36=h2-8h+16+16h2-64h+64

20h=40

h=2

k=4h-3=5

radius=\sqrt{\left ( 2-2 \right )^{2}+\left ( 3-5 \right )^{2}}=2

Therefore equation of the circle is : (x-2)2+(y-5)2=4

x2+y2-4x-10y+25=0

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question