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12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3).

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As we know,

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

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

So let the circle be,

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

Since it's radius is 5 and its centre lies on x-axis,

$(x-h)^2+(y-0)^2=5^2$

And Since it passes through the point (2,3).

$(2-h)^2+(3-0)^2=5^2$

$(2-h)^2=25-9$

$(2-h)^2=16$

$(2-h)=4\:or\:(2-h)=-4$

$h=-2\: or\;6$

When $h=-2\:$ ,The equation of the circle is :

$(x-(-2))^2+(y-0)^2=5^2$

$x^2+4x+4+y^2=25$

$x^2+y^2+4x-21=0$

When $h=6$ The equation of the circle is :

$(x-6)^2+(y-0)^2=5^2$

$x^2-12x+36+y^2=25$

$x^2+y^2-12x+11=0$

Posted by

Pankaj Sanodiya

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