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Questions  >>  CBSE XI  >>  Math  >>  Conic Sections
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Q)

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

$\begin{array}{1 1}x^2+y^2+4x-10y+25=0\\x^2+y^2-4x-10y-38=0\\x^2+y^2-4x-10y+38=0\\x^2+y^2+4x+10y+25=0\end{array} $

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A)
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  • General equation of the circle is x2+y2+2gx+2fy+c=0" role="presentation" style="position: relative;">x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0 where g" role="presentation" style="position: relative;">ggg and f" role="presentation" style="position: relative;">fff are the coordinates of the circle.
Answer : x2+y2−4x−10y+25=0" role="presentation" style="position: relative;">x2+y24x10y+25=0x2+y2−4x−10y+25=0x^2+y^2-4x-10y+25=0
It is given that the circle passes through the points (2,3)" role="presentation" style="position: relative;">(2,3)(2,3)(2,3) and (4,5)" role="presentation" style="position: relative;">(4,5)(4,5)(4,5)
Now substituting these values for x" role="presentation" style="position: relative;">xxx and y" role="presentation" style="position: relative;">yyy in the general equation of the circle we get
x2+y2+2gx+2fy+c=0" role="presentation" style="position: relative;">x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0
(2)2+(3)2+2(2)g+2(3)f+c=0" role="presentation" style="position: relative;">(2)2+(3)2+2(2)g+2(3)f+c=0(2)2+(3)2+2(2)g+2(3)f+c=0(2)^2+(3)^2+2(2)g+2(3)f+c=0
⇒4g+6f+c=−13" role="presentation" style="position: relative;">4g+6f+c=13⇒4g+6f+c=−13\Rightarrow 4g+6f+c=-13------(1)
(4)2+(5)2+2(4)g+2(5)f+c=0" role="presentation" style="position: relative;">(4)2+(5)2+2(4)g+2(5)f+c=0(4)2+(5)2+2(4)g+2(5)f+c=0(4)^2+(5)^2+2(4)g+2(5)f+c=0
16+25+8g+10f+c=0" role="presentation" style="position: relative;">16+25+8g+10f+c=016+25+8g+10f+c=016+25+8g+10f+c=0
⇒8g+10f+c=−41" role="presentation" style="position: relative;">8g+10f+c=41⇒8g+10f+c=−41\Rightarrow 8g+10f+c=-41-----(2)
It is also given the centre of the circle lies on the line y−4x−3=0" role="presentation" style="position: relative;">y4x3=0y−4x−3=0y-4x-3=0
⇒f−4g+3=0" role="presentation" style="position: relative;">f4g+3=0⇒f−4g+3=0\Rightarrow f-4g+3=0
Or −4g+f=3" role="presentation" style="position: relative;">4g+f=3−4g+f=3-4g+f=3-----(3)
Now solving the three equations for g,f" role="presentation" style="position: relative;">g,fg,fg,f and c" role="presentation" style="position: relative;">ccc we get,
4g+6f+c=−13" role="presentation" style="position: relative;">4g+6f+c=134g+6f+c=−134g+\;6f+c=-13
8g+10f+c=−41" role="presentation" style="position: relative;">8g+10f+c=418g+10f+c=−418g+10f+c=-41
on subtracting we get,
−4g−4f=28" role="presentation" style="position: relative;">4g4f=28−4g−4f=28-4g-4f=28
−4g+f=3" role="presentation" style="position: relative;">4g+f=3−4g+f=3-4g+\;f=\;\;3
on subtracting we get
−5f=25" role="presentation" style="position: relative;">5f=25−5f=25-5f=25
f=−5" role="presentation" style="position: relative;">f=5f=−5f=-5
−4g−5=3" role="presentation" style="position: relative;">4g5=3−4g−5=3-4g-5=3
−4g=8" role="presentation" style="position: relative;">4g=8−4g=8-4g=8
⇒g=−2" role="presentation" style="position: relative;">g=2⇒g=−2\Rightarrow g=-2
Substituting for g and f in eq(1)
4(−2)+6(−5)+c=0" role="presentation" style="position: relative;">4(2)+6(5)+c=04(−2)+6(−5)+c=04(-2)+6(-5)+c=0
=>−8−30+c=0" role="presentation" style="position: relative;">=>830+c=0=>−8−30+c=0=> -8-30+c=0
c=38" role="presentation" style="position: relative;">c=38c=38c=38
Hence the equation of the circle is x2+y2−4x−10y+38=0" role="presentation" style="position: relative;">x2+y24x10y+38=0x2+y2−4x−10y+38=0x^2+y^2-4x-10y+38=0
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