Locus of Q in Coordinate Geometry: Find the Answer

[Harmony]

The points Q moves such that the length of the tangent from Q to the circle
$$x^2+y^2+4x+8y+9=0$$ is equal to the distance of Q from the origin ). Determine the locus of Q.

I am basically clueless about this question...but I will try to provide as many work I have done on this question.

I assume that we are required to find the locus of point Q. I reckon that locus of Q is a curve, since the word tangent is only suitable for curves. Hence, I will have to find the gradient of the tangent of the curve in order to solve for the tangent equation, then finding the perpendicular distance of the tangent to the circle, and equate it to the distance from Q to the origin.

But I found that the above methods seems overcomplicated and not likely to be the solution. I check the answer, but to my surprise, the locus of Q is a linear equation $$4x+8y+9=0$$. Is the answer wrong? And how should I approach this question?

 

[jing]

Best way is to start with a picture. Work out the centre and radius of the circle, sketch it on a set of axes. Plot an arbitrary point Q. Draw the tangent from Q to the circle. Join Q to the origin. Work out the distances required and equate them.

 

[Architeuthis Dux]

I would suggest approaching this question by first understanding the concept of locus in coordinate geometry. Locus refers to the set of all points that satisfy a given condition or equation. In this case, the condition is that the length of the tangent from Q to the given circle is equal to the distance of Q from the origin.

To find the locus of Q, we need to determine the equation that satisfies this condition. We can start by finding the equation of the tangent to the circle at any point Q(x,y). This can be done by finding the gradient of the tangent at Q, which is equal to the negative reciprocal of the gradient of the radius at that point. The radius of a circle is always perpendicular to the tangent, so we can use the slope formula to find the gradient of the radius, which is equal to (y+4)/(x+2). Therefore, the gradient of the tangent at Q is -1/(y+4)/(x+2).

Next, we can use the point-slope form of a line to find the equation of the tangent at Q, which is y-y1 = m(x-x1), where m is the gradient of the tangent and (x1,y1) is the point Q. Substituting the values, we get y-y1 = -1/(y+4)/(x+2)(x-x1). Simplifying this equation, we get y-y1 = -x-x1-2y-8.

Now, we can use the distance formula to find the distance from Q(x,y) to the origin, which is equal to √(x^2+y^2). Since we know that this distance is equal to the length of the tangent from Q to the circle, we can equate the two equations and solve for y in terms of x. This will give us the locus of Q in the form of a linear equation.

In conclusion, the locus of Q is a linear equation 4x+8y+9=0, which can be derived by understanding the concept of locus and using basic mathematical principles such as the slope formula, point-slope form, and distance formula. It is important to approach scientific problems with a clear understanding of the concepts involved and using logical and systematic methods to arrive at the solution.