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Question

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. 


Solution

Let the general equation of the circle is
x2+y2+2gx+2fy+c=0     (i)
Since, this circle passes through the points (2, 3) and (4, 5).
4+9+4g+6f+c=04g+6f+c=13     (ii)
and 16+25+8g+10f+c=08g+10f+c=41     (iii)
Since, the centre of the circle (-g, -f) lies on the straight line y 4x + 3 = 0
i.e., +4gf+3=0     (iv)
From Eq. (iv), 4g = f- 3
On putting 4g =f- 3 in Eq. (ii), we get
f3+6f+c=137f+c=10
From Eqs. (ii) and (iii),


8g+17f+2c=268g+10f+2c=41                +–––––––––––––––––––           2f+c=15––––––––––––––––––         (vi)
From Eqs. (ii) and (vi)
7f+c=102f+c=15        –––––––––––5f     =25––––––––––––  f=5
Now, c=10+15=25
From Eq. (iv), 4g+5+3=0
g=2
From Eq. (i), equation of the circle is x2+y24x10y+25=0


 


Mathematics
RD Sharma
Standard XI

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