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The equation of the circle of radius 5 i...

The equation of the circle of radius 5 in the first quadrant which touches the x-axis and the line `3x-4y=0` is `x^2+y^2-24 x-y-25=0` `x^2+y^2-30 x-10 y+225=0` `x^2+y^2-16 x-18 y-64=0` `x^2+y^2-20 x-12 y+144=0`

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Equation of the circle concentric with x^(2)+y^(2)-8x-16y+4=0 and touches y- axis is

Find the equation of the circle which touches the lines 4x-3y+10=0 and 4x-3y-30=0 and whose centre lies on the line 2x+y=0 .

The equation of the circle through the points of intersection of x y -1 -0, x y -2x-4 y l 0 and touching the line x 2y 0, is (B) x y 2 x+20 (C) x y (D) 2 (x2 y x 2

Find the equation of the circle of minimum radius which contains the three cricles x^(2)-y^(2)-4y-5=0 x^(2)+y^(2)+12x+4y+31=0 and x^(2)+y^(2)+6x+12y+36=0

The equation of the circle having the intercept on the line y+2x=0 by the circle x^2 + y^2 + 4x + 6y = 0 as a diameter is : (A) 5x^2 + 5y^2 - 8x + 16y =0 (B) 5x^2 + 5y^2 + 8x - 16y =0 (C) 5x^2 + 5y^2 - 8x - 16y =0 (D) 5x^2 + 5y^2 + 8x+- 16y =0

The equation of a circle with radius 5 and touching both the coordinate axes is x^(2)+y^(2)+-10x+-10y+5=0x^(2)+y^(2)+-10x+-10y=0x^(2)+y^(2)+-10x+-10y+25=0x^(2)+y^(2)+-10x+-10y+51=0

The equation of the circle concentric with the circle x^(2) + y^(2) - 6x - 4y - 12 =0 and touching y axis

The equation of the circle touching Y-axis at (0,3) and making intercept of 8 units on the axis (a) x^(2)+y^(2)-10x-6y-9=0 (b) x^(2)+y^(2)-10x-6y+9=0 (c) x^(2)+y^(2)+10x-6y-9=0 (d) x^(2)+y^(2)+10x+6y+9=0

From a point R(5,8), two tangents RPandRQ are drawn to a given circle S=0 whose radius is 5. If the circumcenter of triangle PQR is (2,3), then the equation of the circle S=0 is x^(2)+y^(2)+2x+4y-10=0x^(2)+y^(2)+x+2y-10=0x^(2)+y^(2)-x+2y-20=0x^(2)+y^(2)+4x-6y-12=0

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