  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=0⇒4g+6f+c=−13     …(ii) and 16+25+8g+10f+c=0⇒8g+10f+c=−41     …(iii) Since, the centre of the circle (-g, -f) lies on the straight line y 4x + 3 = 0 i.e., +4g−f+3=0     …(iv) From Eq. (iv), 4g = f- 3 On putting 4g =f- 3 in Eq. (ii), we get f−3+6f+c=−13⇒7f+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+y2−4x−10y+25=0   MathematicsRD SharmaStandard XI

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