Question

Find the eq of the circle passing through points (4, 1) and (6, 5) whose centre is on line 4x+3y-24=0​

Answer 1

The circle is $(x-p)^2+(y-q)^2=r^2$

Any two points passing the circle will satisfy the above.

So x=4, y=1

$(4-p)^2+(1-q)^2=r^2$

And x=6, y=5

$(6-p)^2+(5-q)^2=r^2$

By subtraction

$p^2-8p+16+q^2-2q+1-(p^2-12p+36+q^2-10q+25)=0$

$-8p+16-2q+1+12p-36+10q-25=0$

$4p+8q-44=0$

∴$p+2q-11=0$

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Since the center (p, q) is on the line,

x=p, y=q will satisfy the linear equation.

So,

∴$4p+3q-24=0$

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∴The center is (p, q)=(3, 4)

As we don't have r value we have to solve once more.

$(x-3)^2+(y-4)^2=r^2$

Since (4, 1) passes the above,

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

∴Which shows $r^2=10$

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Therefore, the circle is $(x-3)^2+(y-4)^2=10$