If x = -2 + 3 cosθ ; y = 1 + 3 sinθ then the locus of the point (x,y) is a circle with
- centre at (-2,1) and radius = 3
- centre at (2,1) and radius = 3
- centre at (2,-1) and radius = 9
- centre at (-2,1) and radius = 9
The Correct Option is A
Solution and Explanation
Top Questions on circle
Let circle \( C \) be the image of
\[ x^2 + y^2 - 2x + 4y - 4 = 0 \]
in the line
\[ 2x - 3y + 5 = 0 \]
and \( A \) be the point on \( C \) such that \( OA \) is parallel to the x-axis and \( A \) lies on the right-hand side of the centre \( O \) of \( C \).
If \( B(\alpha, \beta) \), with \( \beta < 4 \), lies on \( C \) such that the length of the arc \( AB \) is \( \frac{1}{6} \) of the perimeter of \( C \), then \( \beta - \sqrt{3}\alpha \) is equal to:
- From a point A(0,3) on the circle \[ (x + 2)^2 + (y - 3)^2 = 4 \] a chord AB is drawn and extended to a point Q such that AQ = 2AB. Then the locus of Q is:
- A(3,2,0), B(5,3,2), C(-9,6,-3) are three points forming a triangle. AD, the bisector of angle $BAC$ meets BC in D. Find the coordinates of D:
- If \( p \) and \( q \) be the longest and the shortest distance respectively of the point (-7,2) from any point (\(\alpha, \beta\)) on the curve whose equation is \[ x^2 + y^2 - 10x - 14y - 51 = 0 \] then the geometric mean (G.M.) of \( p \) is:
- The locus of the mid-point of a chord of the circle $x^2 + y^2 = 4$ which subtends a right angle at the origin is:
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