Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
AP EAMCET201922 Apr 2019Morning ShiftMathematicsCircleActual

The equation of a circle passing through the point ((2,8) ), touching the lines (4 x-3 y-24=0 ) and (4 x+3 y-42=0 ) and having the (x ) coordinate of its centre less than or equal to 8 is

Options

  1. A(x^2+y^2+2 x-8 y-8=0 )
  2. B(x^2+y^2-4 x-6 y-12=0 )
  3. C(x^2+y^2+4 x-10 y+4=0 )
  4. D(x^2+y^2-6 x-4 y-24=0 )

Correct answer

B. (x^2+y^2-4 x-6 y-12=0 )

Step-by-step solution

By the property of distance, ( aligned & | 4 h-3 k-24 5 |= | 4 h+3 k-42 5 | & = (h-2)^2+(k-8)^2 & 4 h-3 k-24= (4 h+3 k-42) & either 4 h-3 k-24=4 h+3 k-42 & 6 k=18 k=3 & or 4 h-3 k-24=-4 h-3 k+42 & 8 h=66 & h= 66 8 = 33 4 > 8 aligned ) Now, ((4 h-3 k-24)^2=25 [(h-2)^2+(k-8)^2 ] ) By solving this, we get, (h=2 ) ( ) Centre is ((2,3) ). Now, required circle is ( aligned & (x-2)^2+(y-3)^2 =25 & x^2+4-4 x+y^2+9-6 y =25 & x^2+y^2-4 x-6 x =12 aligned )

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