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Question: If a circle of constant radius 3k passes through the origin 0 and meets the coordinate axes at A and...

If a circle of constant radius 3k passes through the origin 0 and meets the coordinate axes at A and B, then the locus of the centroid of triangle OAB is
1. x2+y2=(2k)2{x^2} + {y^2} = {(2k)^2}
2. x2+y2=(3k)2{x^2} + {y^2} = {(3k)^2}
3. x2+y2=(4k)2{x^2} + {y^2} = {(4k)^2}
4. x2+y2=(6k)2{x^2} + {y^2} = {(6k)^2}

Explanation

Solution

Hint: Consider the general equation of circle i.e.x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0 as the equation of the circle of constant radius 3k and also use the formula of radius of circle i.e. r=g2+f2−cr = \sqrt {{g^2} + {f^2} - c} and find the value of x and y to find out the locus of the centroid of triangle OAB.

Complete step-by-step answer:
Let assume x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0be the equation of a circle of constant radius 3k passing through the origin
Since circle is passing through the origin i.e. (x, y) = (0, 0)
Putting the value of x and y in equation of circle
⇒\Rightarrowc = 0
Since the radius of circle is constant
By the formula of radius of a circle r=g2+f2−cr = \sqrt {{g^2} + {f^2} - c}
Since c = 0
Therefore r=g2+f2r = \sqrt {{g^2} + {f^2}}
Putting the value of r and then squaring both sides
\Rightarrow $$$9{k^2} = {g^2} + {f^2}$ (Equation 1) According to the question circles cuts a coordinate axes at A and B Let’s first check the intersection point of circle in the X-axes i.e. (x, 0) \Rightarrow {x^2} + 0 + 2gx + 2f \times 0 + c = 0$ $$ \Rightarrow {x^2} + 2gx = 0 $$ \Rightarrow $$$x(x + 2g) = 0
Since x can’t be 0
Therefore x = - 2g
Let assume that the A is the intersection point in the X-axes
Therefore A = (- 2g, 0)
Let’s check the intersection point of circle in the Y-axes i.e. (0, y)
\Rightarrow $$$0 + {y^2} + 2g \times 0 + 2fy + c = 0$ \Rightarrow {y^2} + 2fy = 0$ $$ \Rightarrow y(y + 2f) = 0Sinceycan’tbe0Thereforey=−2fLetassumethattheBistheintersectionpointintheY−axesThereforeB=(0,−2f)A=(−2g,0)B=(0,−2f)O=(0,0)LetcentroidoftriangleOABisG= Since y can’t be 0 Therefore y = -2f Let assume that the B is the intersection point in the Y-axes Therefore B = (0, -2f) A = (- 2g, 0) B = (0, -2f) O = (0, 0) Let centroid of triangle OAB is G=(h',k')Sinceweknowthatforx= Since we know that for x =\dfrac{{{x_1} + {x_2} + {x_3}}}{3} $$h' = \dfrac{{ - 2g + 0 + 0}}{3}$$ h' = \dfrac{{ - 2g}}{3} g = \dfrac{{ - 3h'}}{2} And for y =$$\dfrac{{{y_1} + {y_2} + {y_3}}}{3}$$ $$k' = \dfrac{{0 - 2f + 0}}{3}$$ k' = \dfrac{{ - 2f}}{3} f = \dfrac{{ - 3k'}}{2} Putting the values of f and g in equation 1 $$ \Rightarrow $$$9{k^2} = {(\dfrac{{ - 3h'}}{2})^2} + {(\dfrac{{ - 3k'}}{2})^2}
\Rightarrow $$$9{k^2} = \dfrac{{9{{h'}^2}}}{2} + \dfrac{{9{{k'}^2}}}{2}$ \Rightarrow $$$4{k^2} = {h'^2} + {k'^2}SinceweassumedthecoordinatesofcentroidoftrianglenowwegotarelationbetweenthemThereforelocusofthecentroidoftriangleOABis Since we assumed the coordinates of centroid of triangle now we got a relation between them Therefore locus of the centroid of triangle OAB is4{k^2} = {x^2} + {y^2}$

Note: In these types of questions taking a general equation of circle since the circle passing through therefore c = 0 then using the formula of radius i.e. (r=g2+f2−cr = \sqrt {{g^2} + {f^2} - c} ) to find the equation and then finding the intersection point in the X and Y axes next assuming that A and B are the intersection point in the X axes and Y axes now assuming the centroid of triangle OAB are (h′,k′)(h',k')now using the formula of x and y coordinates i.e. (x=x1+x2+x33x = \dfrac{{{x_1} + {x_2} + {x_3}}}{3} and y=y1+y2+y33y = \dfrac{{{y_1} + {y_2} + {y_3}}}{3})to find the value of (h′,k′)(h',k')since we got a relation between (h′,k′)(h',k')and (x, y) therefore substituting the x and y by (h′,k′)(h',k')hence we got the locus of the centroid of triangle OAB i.e. 4k2=x2+y24{k^2} = {x^2} + {y^2}