find the equation of circle each satisfying the given cond.

1) center(1,3) and tangent to 2x-y-4=0

2) tangent to 2x-3y-7=0 at (2,-1) and passes through (4,1)

3) tangent ro 3x+y+3=0 at (-3,6) and tangent to x+3y-7

Question

find the equation of circle each satisfying the given cond
1) center(1,3) and tangent to 2x-y-4=0
2) tangent to 2x-3y-7=0 at (2,-1) and passes through (4,1)
3) tangent ro 3x+y+3=0 at (-3,6) and tangent to x+3y-7

Ajanta Trivedi · Accepted Answer

1 center of the circle is (1,3) and the equation of the tangent is 2x-y-4=0 length of the perpendicular from center to the tangent is equal to the radius therefore r = $| \frac{2 * 1 - 1 * 3 - 4}{\sqrt{2^{2} + 1^{2}}} | = | \frac{2 - 3 - 4}{\sqrt{5}} | = | \frac{- 5}{\sqrt{5}} | = \sqrt{5} u n i t s$ the equation of a circle with center (h,k) and radius r is given by: $(x - h)^{2} + (y - k)^{2} = r^{2}$ thus the equation of the circle with center (1,3) and radius $\sqrt{5} u n i t s$ is given by; $(x - 1)^{2} + (y - 3)^{2} = (\sqrt{5})^{2} x^{2} + 1 - 2 x + y^{2} + 9 - 6 y = 5 x^{2} + y^{2} - 2 x - 6 y + 5 = 0$ which is the required equation of the circle (2) let the equation of the tangent at point A (2,-1) is 2x-3y-7=0 and the circle passes through point B (4,1)

the intersection of the normal at point A and the perpendicular bisector of AB is the center let AO be the normal at point A(2,-1) slope of the line AO is $- \frac{1}{2 / 3} = - \frac{3}{2}$ the equation of the line AO is $y + 1 = - \frac{3}{2} (x - 2) 2 y + 2 = - 3 x + 6 3 x + 2 y = 4 (1)$ the perpendicular bisector of the line AB will pass through the mid-point M= $(\frac{2 + 4}{2}, \frac{- 1 + 1}{2}) = (3, 0)$ the slope of AB= $\frac{- 1 - 1}{2 - 4} = \frac{- 2}{- 2} = 1$ the equation of the perpendicular bisector of AB is $y - 0 = - 1 (x - 3) x + y = 3$ (2) solving (1) and (2), 3x+2y=4 2x+2y=6 x= -2 and y=5 therefore the center of the circle is O(-2,5) and radius $r^{2} = O A^{2} = (- 2 - 2)^{2} + (5 + 1)^{2} = 16 + 36 = 52$ thus the equation of the circle is $(x + 2)^{2} + (y - 5)^{2} = 52 x^{2} + 4 + 4 x + y^{2} + 25 - 10 y = 52 x^{2} + y^{2} + 4 x - 10 y - 23 = 0$ which is the required equation of the circle for 2nd ques 3

let (h, k) be the center of the circle the equation of the normal at point A is given by $y - 6 = - \frac{1}{- 3} (x + 3) y - 6 = \frac{1}{3} (x + 3) 3 y - 18 = x + 3 x - 3 y = - 21$ since the center will pass through this line h-3k=-21 (1) since x+3y-7=0 is also the tangent to the circle since the lengths of the perpendicular from O to the tangents are equal to radius $| \frac{3 h + k + 3}{\sqrt{3^{2} + 1}} | = | \frac{h + 3 k - 7}{\sqrt{1 + 3^{2}}} | \Rightarrow | 3 h + k + 3 | = | h + 3 k - 7 |$ $3 h + k + 3 = \pm (h + 3 k - 7) 3 h + k + 3 = h + 3 k - 7 2 h - 2 k + 10 = 0 h - k = - 5 (2) 3 h + k + 3 = - (h + 3 k - 7) 4 h + 4 k = 4 h + k = 1 (3)$ solving (1) and (2) h-3k=-21 h-k=-5 -2k=-16 k=16/2=8 h=-5+8=3 therefore the center of the circle is (3,8) radius OA= $\sqrt{(- 3 - 3)^{2} + (6 - 8)^{2}} = \sqrt{6^{2} + 2^{2}} = \sqrt{36 + 4} = \sqrt{40}$ the equation of the circle is $(x - 3)^{2} + (y - 8)^{2} = 40 x^{2} - 6 x + 9 + y^{2} - 16 y + 64 = 40 x^{2} + y^{2} - 6 x - 16 y + 33 = 0$ (4) solving (1) and (3) h-3k=-21 h+k=1 -4k=-22 k=22/4=11/2=5 5 h=1-5 5=-4 5 the center of the circle is (-4 5,5 5) the radius of the circle is $\sqrt{(- 4 5 + 3)^{2} + (5 5 - 6)^{2}} = \sqrt{{1 5}^{2} + (0 5)^{2}} = \sqrt{2 25 + 25} = \sqrt{2 5}$ thus the equation of the circle is $(x + 4 5)^{2} + (y - 5 5)^{2} = 2 5 x^{2} + 9 x + 20 25 + y^{2} - 11 y + 30 25 = 2 5 x^{2} + y^{2} + 9 x - 11 y + 48 = 0$ (5) thus the equations of the required circles are eq (4) and (5) hope this helps you cheers!!

find the equation of circle each satisfying the given cond. 1) center(1,3) and tangent to 2x-y-4=0 2) tangent to 2x-3y-7=0 at (2,-1) and passes through (4,1) 3) tangent ro 3x+y+3=0 at (-3,6) and tangent to x+3y-7

find the equation of circle each satisfying the given cond.

1) center(1,3) and tangent to 2x-y-4=0

2) tangent to 2x-3y-7=0 at (2,-1) and passes through (4,1)

3) tangent ro 3x+y+3=0 at (-3,6) and tangent to x+3y-7