Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
AP EAMCET201823 Apr 2018Evening ShiftMathematicsCircleActual

If a circle touches the lines 3 x-4 y-10=0 and 3 x-4 y+30=0 and its centre lies on the line x+2 y=0 , then the equation of the circle is

Options

  1. Ax^2+y^2+4 x-2 y-11=0
  2. Bx^2+y^2+2 x-4 y-11=0
  3. Cx^2+y^2-4 x+2 y-11=0
  4. Dx^2+y^2+2 x-y-11=0

Correct answer

A. x^2+y^2+4 x-2 y-11=0

Step-by-step solution

Distance between parallel lines 3 x-4 y-10=0 and 3 x-4 y+30=0 is the length of diametre of required circle, so radius = 1 2 40 9+16 =4 and mid-point of intersection of lines 3 x-4 y-10=0, x+2 y=0 and 3 x-4 y+30=0 , x+2 y=0 , is the centre of required circle, so centre is ( 2-6 2 , -1+3 2 )=(-2,1) , then equation of required circle is, x^2+y^2+4 x-2 y-11=0 .

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 AP EAMCET PYQs