JEE MAIN - Mathematics (2015 (Offline) - No. 15)
Locus of the image of the point $$(2, 3)$$ in the line $$\left( {2x - 3y + 4} \right) + k\left( {x - 2y + 3} \right) = 0,\,k \in R,$$ is a :
circle of radius $$\sqrt 2 $$.
circle of radius $$\sqrt 3 $$.
straight line parallel to $$x$$-axis
straight line parallel to $$y$$-axis
Explanation
Intersection point of $$2x - 3y + 4 = 0$$
and $$x-2y+3=0$$ is $$(1, 2)$$
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Since, $$P$$ is the fixed point for given family of lines
So, $$PB=PA$$
$${\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = {\left( {2 - 1} \right)^2} + {\left( {3 - 2} \right)^2}$$
$${\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = 1 + 1 = 2$$
$${\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {\sqrt 2 } \right)^2}$$
$${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$$
Therefore, given locus is a circle with center $$(1, 2)$$ and radius $$\sqrt 2 .$$
and $$x-2y+3=0$$ is $$(1, 2)$$
Since, $$P$$ is the fixed point for given family of lines
So, $$PB=PA$$
$${\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = {\left( {2 - 1} \right)^2} + {\left( {3 - 2} \right)^2}$$
$${\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = 1 + 1 = 2$$
$${\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {\sqrt 2 } \right)^2}$$
$${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$$
Therefore, given locus is a circle with center $$(1, 2)$$ and radius $$\sqrt 2 .$$
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