The locus of a point which moves in a plane so that the sum of the squares of its distances from the lines $a x + b y + c = 0$ and $b x - a y + d = 0$ is $r^{2}$ , is a circle of radius

r \sqrt{(a^{2} + b^{2})}

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r

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r \sqrt{a^{2} - b^{2}}

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2 r

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Solution

The correct option is A $r$
The given lines are perpendicular.
So shifting the origin to their point of intersection
If $(h, k)$ are the new coordinates of any point on the locus
Then $h^{2} + k^{2} = r^{2}$
and the locus is $x^{2} + y^{2} = r^{2}$
Which is circle of radius $r$

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The locus of a point which moves in a plane so that the sum of the squares of its distances from the lines ax+by+c=0 and bx−ay+d=0 is r2, is a circle of radius

The correct option is A rThe given lines are perpendicular.So shifting the origin to their point of intersection If (h,k) are the new coordinates of any point on the locus Then h2+k2=r2 and the locus is x2+y2=r2Which is circle of radius r

The locus of a point which moves in a plane so that the sum of the squares of its distances from the lines $a x + b y + c = 0$ and $b x - a y + d = 0$ is $r^{2}$ , is a circle of radius

The correct option is A $r$
The given lines are perpendicular.
So shifting the origin to their point of intersection
If $(h, k)$ are the new coordinates of any point on the locus
Then $h^{2} + k^{2} = r^{2}$
and the locus is $x^{2} + y^{2} = r^{2}$
Which is circle of radius $r$