Question
The two points A and B in a plane are such that for all points P lies on circle satisfied
, then k will not be equal to
- 0
- 1
- 2
- None of these
The correct answer is: 1
The locus is the collection of all points that satisfy the requirements and create geometrical shapes like lines, line segments, circles, curves, etc. Only curved shapes are defined for the locus. Both regular and irregular shapes are possible.
The circle's equation with its centre at (h, k) and radius 'a' is (x-h)2+(y-k)2 = a2 which is called the standard form for the equation of a circle.
Now we have given the two points A and B in a plane are such that for all points P lies on circle satisfied
.
Consider the circle as: x2+y2=1.
Now lets assume the points to be A(1,0), B(0,1), P(cosθ, sinθ).
Now we get:
![fraction numerator P A over denominator P B end fraction equals fraction numerator square root of left parenthesis 1 minus cos theta right parenthesis squared plus sin to the power of 2 end exponent theta end root over denominator square root of cos squared theta plus left parenthesis 1 minus sin theta right parenthesis squared end root end fraction
S o l v i n g space t h i s comma space w e space g e t colon
fraction numerator P A over denominator P B end fraction equals fraction numerator square root of left parenthesis 1 minus 2 cos theta plus cos squared theta right parenthesis plus sin squared theta end root over denominator square root of cos squared theta plus left parenthesis 1 minus 2 sin theta plus sin squared theta right parenthesis end root end fraction
N o w space w e space k n o w colon space sin squared theta plus cos squared theta space equals space 1 comma space s o space p u t t i n g space t h i s comma space w e space g e t colon
fraction numerator P A over denominator P B end fraction equals fraction numerator square root of 1 minus 2 cos theta plus 1 end root over denominator square root of 1 plus 1 minus 2 sin theta end root end fraction
fraction numerator P A over denominator P B end fraction equals fraction numerator square root of 2 minus 2 cos theta end root over denominator square root of 2 minus 2 sin theta end root end fraction
fraction numerator P A over denominator P B end fraction equals fraction numerator square root of 1 minus 1 cos theta end root over denominator square root of 1 minus 1 sin theta end root end fraction equals k
i f space k equals 0 comma space t h e n space cos theta equals 1 comma space p o s s i b l e
i f space k equals 1 comma space t h e n space sin theta equals cos theta comma space p o s s i b l e
i f space k equals 2 comma space t h e n space 1 minus cos theta equals 4 minus 4 sin theta
S o colon
sin left parenthesis theta minus alpha right parenthesis equals 3 square root of 17 space p o s s i b l e
S o space t h e space o p t i o n space i s space n o t space p o s s i b l 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)
Here we have given the two points A and B in a plane are such that for all points P lies on circle satisfied, so first we will get the equation of circle. Then we will find value of k with assumptions of trigonometric functions and then will find the final answer.
So here we have given the two points A and B in a plane are such that for all points P lies on circle satisfied . We can also use equation of circle to find the answer. So here the correct option is none of these.
Related Questions to study
A circle touches the x-axis and also touches the circle with centre at (0, 3) and radius. The locus of the centre of the circle is
So here we have given to find the locus is in which form. We used the concept of circles and solved the question. We can also use equation of circle to find the answer. So the locus is x2-10y+5=0, this is the representation of parabola, so correct answer is parabola.
A circle touches the x-axis and also touches the circle with centre at (0, 3) and radius. The locus of the centre of the circle is
So here we have given to find the locus is in which form. We used the concept of circles and solved the question. We can also use equation of circle to find the answer. So the locus is x2-10y+5=0, this is the representation of parabola, so correct answer is parabola.
If d is the distance between the centres of two circles,
are their radii and
, then
For such questions, we have to see how the cricles are with respect to each other to find the distance between the centers.
If d is the distance between the centres of two circles,
are their radii and
, then
For such questions, we have to see how the cricles are with respect to each other to find the distance between the centers.