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11. Find the equation of the circle passing through the points (2,3) and (–1,1) and hose centre is on the line $x - 3y - 11 = 0$ .

Answers (1)

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As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given Here,

Condition 1: the circle passes through points (2,3) and (–1,1)

$(2-h)^2+(3-k)^2=r^2$

$(-1-h)^2+(1-k)^2=r^2$

Here,

$(2-h)^2+(3-k)^2=(-1-h)^2+(1-k)^2$

$(2-h)^2-(-1-h)^2+(3-k)^2-(1-k)^2=0$

$(3)(1-2h)+(2)(4-2k)=0$

$3-6h+8-4k=0$

$6h+4k=11$

Now, Condition 2: centre is on the line. $x - 3y - 11 = 0$

$h-3k=11$

From condition 1 and condition 2

$h=\frac{7}{2},\:k=\frac{-5}{2}$

Now let's substitute this value of h and k in condition 1 to find out $r$

$\left ( 2-\frac{7}{2}\right )^2+\left (3+\frac{5}{2}\right )^2=r^2$

$\frac{9}{4}+\frac{121}{4}=r^2$

$r^2=\frac{130}{4}$

So now, the Final Equation of the circle is

$\left(x-\frac{7}{2}\right )^2+\left(y+\frac{5}{2}\right)^2=\frac{130}{4}$

$x^2-7x+\frac{49}{4}+y^2+5y+\frac{25}{4}=\frac{130}{4}$

$x^2+y^2-7x+5y-\frac{56}{4}=0$

$x^2+y^2-7x+5y-14=0$

Posted by

Pankaj Sanodiya

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