A circle passes through the point (3, 4) and | vedclass

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(B) Let the circle be $(x - h)^2 + (y - k)^2 = r^2$,where $(h, k)$ is the centre.Since it passes through $(3, 4)$,we have $(3 - h)^2 + (4 - k)^2 = r^2

Vedclass · Accepted Answer

$225$

Vedclass · Answer

(B) Let the circle be $(x - h)^2 + (y - k)^2 = r^2$,where $(h, k)$ is the centre.Since it passes through $(3, 4)$,we have $(3 - h)^2 + (4 - k)^2 = r^2$,which implies $h^2 - 6h + 9 + k^2 - 8k + 16 = r^2$,or $h^2 + k^2 - 6h - 8k + 25 = r^2$.The equation of the circle is $x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0$.Substituting $r^2 = h^2 + k^2 - 6h - 8k + 25$,the equation becomes $x^2 + y^2 - 2hx - 2ky + 6h + 8k - 25 = 0$.This circle cuts $x^2 + y^2 - a^2 = 0$ orthogonally,so $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Here $g_1 = -h, f_1 = -k, c_1 = 6h + 8k - 25$ and $g_2 = 0, f_2 = 0, c_2 = -a^2$.Thus,$2(-h)(0) + 2(-k)(0) = 6h + 8k - 25 - a^2$.This gives $6h + 8k - 25 - a^2 = 0$.The locus of the centre $(h, k)$ is the line $6x + 8y - (25 + a^2) = 0$.The distance of this line from $(0, 0)$ is $\frac{|-(25 + a^2)|}{\sqrt{6^2 + 8^2}} = 25$.$\frac{25 + a^2}{10} = 25$ $\Rightarrow 25 + a^2 = 250$ $\Rightarrow a^2 = 225$.

$A$ circle passes through the point $(3, 4)$ and cuts the circle $x^2 + y^2 = a^2$ orthogonally. The locus of its centre is a straight line. If the distance of this straight line from the origin is $25$,then $a^2$ is equal to:

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