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10. Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line $4x + y = 16$ .

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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 (4,1) and (6,5)

$(4-h)^2+(1-k)^2=r^2$

$(6-h)^2+(5-k)^2=r^2$

Here,

$(4-h)^2+(1-k)^2=(6-h)^2+(5-k)^2$

$(4-h)^2-(6-h)^2+(1-k)^2-(5-k)^2=0$

$(-2)(10-2h)+(-4)(6-2k)=0$

$-20+4h-24+8k=0$

$4h+8k=44$

Now, condition 2: centre is on the line $4x + y = 16$ .

$4h+k=16$

From condition 1 and condition 2

$h=3,\:k=4$

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

$(4-3)^2+(1-4)^2=r^2$

$1+9=r^2$

$r=\sqrt{10}$

So now, the final equation of the circle is

$(x-3)^2+(y-4)^2=(\sqrt{10})^2$

$x^2-6x+9+y^2-8y+16=10$

$x^2+y^2-6x-8y+15=0$

Posted by

Pankaj Sanodiya

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