Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
KCET2010MathematicsCircle

Equation of the circle centred at (4,3) touching the circle x²+y²=1 externally, is

Options

  1. Ax²+y²-8 x-6 y+9=0
  2. Bx²+y²+8 x+6 y+9=0
  3. Cx²+y²+8 x-6 y+9=0
  4. Dx²+y²-8 x+6 y+9=0

Correct answer

A. x²+y²-8 x-6 y+9=0

Step-by-step solution

Given that, equation of circle x²+y²=1 Centre at O (0,0) Radius = OA =1 Also, the centre of another circle C(4,3) both circle touch externally. Then, distance between centres = OC. = (4-0)²+(3-0)² = 16-9 =5 Now, AC = OC - OA A C=5-1=4 So, the radius of other circle is 4 . Now, the equation of other circle touch externally to the circle x ²+ y ²=1 is, gathered (x-4)²+(y-3)²=16 x²+y²-8 x-6 y+9=0 gathered

Practice Circle on Quantrex Academy →

More from Circle

A circle touches both the coordinate axes and the straight line L 4 x +3 y -6=0 in the first quadrant. If this circle lies below the line L =0 , then the equation of that circle is 2025If the smallest circle through the points of intersection of x^2+y^2=a^2 and x +y =p, 0 < p < a is x^2+y^2-a^2+ (x +y -p)=0 then = 2025If the lines 3 x-4 y+4=0 and 6 x-8 y-7=0 are the tangents to the same circle, then the area of that circle (in sq.units) is 2025Circles are drawn through the point (2,0) to cut intercepts of length 5 units on the X -axis. If their centre lie in the first quadrant, then their equation is 2025The circles x^2+y^2-2 x-4 y-4=0 and x^2+y^2+2 x+4 y-11=0 2025If the line 4 x-3 y+7=0 touches the circle x^2+y^2-6 x+4 y-12=0 at ( , ) , then +2 = 2025The slope of the common tangent drawn to the circles x^2+y^2-4 x+12 y-216=0 and x^2+y^2+6 x-12 y+36=0 is 2025If r₁ and r₂ are radii of two circles touching all the four circles (x r)^2+(y r)^2=r^2 , then r₁+r₂ r = 2025 Full Circle list All KCET PYQs