The line $ax + by + c = 0$ is normal to the circle $x^2 + y^2 + 2gx + 2fy + d = 0$ if
- A$ag + bf + c = 0$
- B$ag + bf - c = 0$
- C$ag - bf + c = 0$
- D$ag - bf - c = 0$
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Similar Questions
Find the equation of the circle having normals $(x-1)(y-2)=0$ and a tangent $3x+4y=6$.
The equation of one of the common tangents of the circle $x^2+y^2-6y+4=0$ and the parabola $y^2=x$ is
Match the points on the curve $2y^2 = x + 1$ with the slopes of the normals at those points and choose the correct answer.
$A. (7, 2)$ $1. -4\sqrt{2}$ $B. (0, 1/\sqrt{2})$ $2. -8$ $C. (1, -1)$ $3. 4$ $D. (3, \sqrt{2})$ $4. 0$ $5. -2\sqrt{2}$
| $A. (7, 2)$ | $1. -4\sqrt{2}$ |
| $B. (0, 1/\sqrt{2})$ | $2. -8$ |
| $C. (1, -1)$ | $3. 4$ |
| $D. (3, \sqrt{2})$ | $4. 0$ |
| $5. -2\sqrt{2}$ |
DifficultTS EAMCET 2004
View SolutionThe equation of a circle which has a tangent $3x + 4y = 6$ and two normals given by $(x - 1)(y - 2) = 0$ is
The equations of the tangents drawn from the origin to the circle $x^2+y^2+2gx+2fy+g^2=0$ are
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