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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the equation of the circle passing through the points $(2,3)$ and $(-1,1)$ and whose centre is on the line $x-3y-11=0$

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  • Given C (h, k) be the centre and r the radius of circle. Let P(x, y) be any point on the circle. Then, by the definition, | CP | = r . By the distance formula, we have, $(x-h)^2+(y-k)^2=r^2$
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  • Now, given two pairs of co-ordinates through which the circle passes and the equation of the line through which the circle passes, we can substitute those values in the equation of the circle and the line and solve for h, k and r.
Given two pairs of co-ordinates through which the circle passes and the equation of the line through which the circle passes, we can substitute those values in the equation of the circle and the line and solve for h, k and r.
Given that the circle passes through $(2,3) \rightarrow (2-h)^2 + (3-k)^2 = r^2... (1)$
Given that the circle passes through $(-1,1) \rightarrow (-1-h)^2 + (1-k)^2 = r^2 ... (2)$
From $(1)$ and $(2)$, we get, $ ((2-h)^2 + (3-k)^2 = (-1-h)^2 + (1-k)^2 \rightarrow 6h + 4k = 11.. (3)$
Since the centre of the circle lines on $x-3y-11=0 \rightarrow h-3k=11... (4)$
Solving $(3)$ and $(4)$, we get, $h = \large\frac{7}{2}$ and $k=\large\frac{-5}{2}$
Substituting the values of $h$ and $k$ in $(1)$, we get: $(2-\large\frac{7}{2})^2$$+(y-\large\frac{-5}{2})^2$$ = r^2 \rightarrow \large\frac{-3}{2}^2 $$ + \large\frac{11}{2}^2$$ = r^2$
$\Rightarrow \large\frac{9}{4}$$ + \large\frac{121}{4}$$ = r^2 $$= \large\frac{65}{2} = r^2$
Now, we can substitute for $h, k$ and $r$ in $(x-h)^2+(y-k)^2=r^2$ to get the equation of the circle:
$\quad \quad (x- \large\frac{7}{2})^2$$ + (y+\large\frac{5}{2})^2$$ = \large\frac{65}{2}$
$\quad \quad 4x^2-28x+49+4y^2+20y+25 = 130 \rightarrow 4x^2+4y^2-28x+20y-56 = 0 $
$\quad \quad x^2+y^2-7x+5y-14 = 0$
answered Apr 1, 2014 by balaji.thirumalai
edited Apr 1, 2014 by balaji.thirumalai
 

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