Question

Let P be a point on the circle x²+y²-6x-12y+40=0 at farthest distance from origin. If it is rotated on circle through an angle of 90° in anticlockwise sense , the coordinates of point in new position would be ?​

Answer 1

Answer:

The points on maximum and minimum distance from (-9,7) must lie on normal from it to the circle.

Circle is (x−3)

2

+(y+2)

2

=25

Centre of circle is (3,−2)

Equation of normal passing through (−9,7) and (3,−2) is 3x+4y−1=0

Intersection of normal and circle gives two points which are (−1,1) and (7,−5).

Clearly, (7,−5) is at maximum distance from (−9,7).

'

Answer 2

Answer:

The coordinates of point in new position would be (4,8).

Step-by-step explanation:

As per the data given in the questions,

We have to calculate coordinates of point in new position.

As per the questions

It is given that

P be a point on the circle x²+y²-6x-12y+40=0 at farthest distance from origin. If it is rotated on circle through an angle of 90° in anticlockwise sense.

Given question of circle is x²+y²-6x-12y+40=0.

If we substitute x=0 and y=0 then 40>0.that is origin is out side the circle.

We know the general formula of equation of circle is $x^{2} +y^{2}+2gx+2fy+c=0$

Compare circle equation and given equation we found

2g=-6⇒g=-3

Similarly 2f=-12⇒f=-6

and c=40

As we know center of circle is O (-g,-f)

So, O (3,6) is center.

Now we will find the fastest point y=mx

= y=$\frac{6-0}{3-0} = 2$x

⇒y=2x

put y=2x in circle equation we get

$x^{2} +4x^{2} -6x-24x+40=0$

⇒$5x^{2} -30x+40=0$

or $x^{2} -6x+8=0$

$=x^{2} -4x-2x+8=0$

= x (x-4)-2(x-4)=0

=(x-2)(x-4)=0

= x= 2, 4

put the value of x in y=2x

then point will be (2,4) and (4,8)

So, fastest distance from origin point is (4,8)

Hence, the coordinates of point in new position would be (4,8).

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