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13. Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.

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Let the equation of circle be,

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

Now since this circle passes through (0,0)

$(0-h)^2+(0-k)^2=r^2$

$h^2+k^2=r^2$

Now, this circle makes an intercept of a and b on the coordinate axes.it means circle passes through the point (a,0) and (0,b)

So,

$(a-h)^2+(0-k)^2=r^2$

$a^2-2ah+h^2+k^2=r^2$

$a^2-2ah=0$

$a(a-2h)=0$

$a=0\:or\:a-2h=0$

Since $a\neq0\:so\:a-2h=0$

$h=\frac{a}{2}$

Similarly,

$(0-h)^2+(b-k)^2=r^2$

$h^2+b^2-2bk+k^2=r^2$

$b^2-2bk=0$

$b(b-2k)=0$

Since b is not equal to zero.

$k=\frac{b}{2}$

So Final equation of the Circle ;

$\left ( x-\frac{a}{2} \right )^2+\left ( y-\frac{b}{}2 \right )^2=\left ( \frac{a}{2} \right )^2+\left ( \frac{b}{2} \right )^2$

$x^2-ax+\frac{a^2}{4}+y^2-bx+\frac{b^2}{4}=\frac{a^2}{4}+\frac{b^2}{4}$

$x^2+y^2-ax-bx=0$

Posted by

Pankaj Sanodiya

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