Equation of a circle which passes through (3,6) and touches the axis is
(a) x² + y2 +6x + 6y +3 = 0
(b) x² + y² - 6x -6y - 9 = 0
(C) x² + y² - 6x - 6y + 9 = 0
(d) x² + y² - 6x + 6y - 3=0
Answers
Answer :
(C) x² + y² - 6x - 6y + 9 = 0
Note :
- The general equation of a circle is given by ; (x - h)² + (y - k)² = r² , where (h,k) is the centre and r is the radius of the circle .
Solution :
Here ,
It is given that required circle passes through the point (3,6) and touches the coordinate axes .
From the given data we can conclude that , the required circle completely lies in 1st quadrant (°•° The point (3,6) lies in 1st quadrant) .
Also ,
Since the required circle lies in 1st quadrant and touches the coordinate axes , thus we can conclude that the center of the circle is (r,r) , where r is the radius of the circle .
Thus ,
The equation of the required circle will be given as ; (x - r)² + (y - r)² = r² -----(1)
Now ,
Since the required circle passes through the point (3,6) , thus x = 3 , y = 6 must satisfy the equation of the circle .
Now ,
Putting x = 3 and y = 6 in eq-(1) , we get ;
=> (3 - r)² + (6 - r)² = r²
=> (3² + r² - 2•3•r) + (6² + r² - 2•6•r) = r²
=> 9 + r² - 6r + 36 + r² - 12r = r²
=> r² - 18r + 45 = 0
=> r² - 15r - 3r + 45 = 0
=> r(r - 15) - 3(r - 15) = 0
=> (r - 15)(r - 3) = 0
=> r = 15 , 3
• If r = 15 , then eq-(1) will become ;
=> (x - 15)² + (y - 15)² = 15²
=> x² + 225 - 30y + y² + 225 - 30y = 225
=> x² + y² - 30x - 30y + 225 = 0
• If r = 3 , then eq-(1) will become ;
=> (x - 3)² + (y - 3)² = 3²
=> x² + 9 - 6x + y² + 9 - 6y = 9
=> x² + y² - 6x - 6y + 9 = 0
Hence ,
Required answer is option(C) .
Answer:
The answer is a) x² + y2 +6x + 6y +3 = 0 is the correct answer ok