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

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(B) Let the circle be $(x-h)^2 + (y-k)^2 = r^2$. Since it passes through $(3, 4)$,we have $(3-h)^2 + (4-k)^2 = r^2$,which simplifies to $h^2 - 6h + 9

Vedclass · Accepted Answer

$225$

Vedclass · Answer

(B) Let the circle be $(x-h)^2 + (y-k)^2 = r^2$. Since it passes through $(3, 4)$,we have $(3-h)^2 + (4-k)^2 = r^2$,which simplifies to $h^2 - 6h + 9 + k^2 - 8k + 16 = r^2$,or $h^2 + k^2 - 6h - 8k + 25 = r^2$. The equation of the circle is $x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0$. Since it cuts $x^2 + y^2 - a^2 = 0$ orthogonally,the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ gives $2(-h)(0) + 2(-k)(0) = (h^2 + k^2 - r^2) - a^2$. Thus,$h^2 + k^2 - r^2 = a^2$. Substituting $r^2 = h^2 + k^2 - a^2$ into the first equation: $h^2 + k^2 - 6h - 8k + 25 = h^2 + k^2 - a^2$. This simplifies to $6h + 8k - 25 - a^2 = 0$. The locus of the centre $(h, k)$ is the line $6x + 8y - (25 + a^2) = 0$. The distance of this line from the origin $(0, 0)$ is given as $25$. So,$\frac{|-(25 + a^2)|}{\sqrt{6^2 + 8^2}} = 25$. $\frac{25 + a^2}{10} = 25$. $25 + a^2 = 250$. $a^2 = 225$.

$A$ 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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