Question

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Answer 1

SOLUTION ::

Let A & B be any two points on the circle

Let R (p, q) be the mid point of the secants intercepted by the circle

Let C be the Centre of the Circle & P be the fixed point (h, k)

Now the Co-ordinate of C is (0,0)

By the given condition

CR is perpendicular to RP

SO Slope of CR × Slope RP = - 1

=> q/p × (k-q) /(h-p) = - 1

=> kq - q² = - hp + p²

=> p² + q² = hp + kq

So the required locus is given by

x² + y² = hx + ky

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Answer 2

Point Q lies on the circle having diameter OP.  

The locus of point Q:  

Let R (p, q) be the mid point of the secants intercepted by the circle

(x - 0)(x - h) + (y - 0)(y - k) = 0  Answer:

Step-by-step explanation:

Let A( x, y) be any point on the Locus.

The given Straight line passes through B (h, k).

Since A, B are both the points on the same line, They are collinear and B is the foot of the perpendicular from O.

So, AB ⊥ AO

x² + y² - hx - ky = 0

Suppose the equation of line passing through point (h,k) is y=mx+c…eqn(1)

Point P is (0,c) and point Q is (  

Relation b/w c and m according to eqn (1) is m=  

h

k−c

For rectangle OPRS coordinates of R will be (  

​

If coordinates of R are (h  

) Then,  

=1hk

$Now the Co-ordinate of C is (0,0)Then CR is perpendicular to RPSO Slope of CR × Slope RP = - 1$

Putting the value of c from eqn(3) in eqn(2), We get  

hsubstituting (h  

with (x,y), We get, =1

Product of slopes of perpendicular lines is - 1

Let P(h, k) be the given point, let Q(x, y) be the foot of the perpendicular, and let O be the origin. The line can have any direction.  

∠PQO = 90°  

Point Q lies on the circle having diameter OP.  

The locus of point Q:  

(x - 0)(x - h) + (y - 0)(y - k) = 0  

x² + y² - hx - ky = 0

This is required locus of foot of perpendicular drawn from origin to a variable straight line passing through (h, k).

So the required locus is given by

$x^2 + y^2 = hx + ky$

Hope it helps