2005 Mocks Circle Question

fearcruach

Has anyone got the answers to this question because i have to do it for homework and i have some messed up answers but i dont know if they are right or not. The answers would be a great help.

Stark

Post the question. Not all mock papers are the same and you didn't even specify if it was ordinary level/higher level (I assume you're on about Maths?)

Gileadi

wouldnt worry about the answer being right once you used the right methods you get most of the marks...

fearcruach

Yeah higher level maths it was. Its just I'm getting two circle questions in the summer exam and I'm having trouble with the (c) parts, so I'm trying desperately to get them right and I don't know if I'm even using the right method. Stupid maths teacher gives us the one year we don't have the answers to.

The question:

1. (a) Find the equation of the circle passing through the points (-3, 0),
(2, 1) and (1, -4).

(b) Find the equation of the tangents from the point (-2, 1) to the circle x2 + y2 - 10x - 4y +19 = 0

(c) Find the radius length and the centres of the circles passing through the point (-2, 4) having the X - axis and the Y - axis as tangents.

(a) Many ways to do this.

Put all 3 points into the equation x^2 + y^2 +2gx + 2fy + c=0.

You know have 3 unknowns and 3 equations. Find f,g and c by solving the 3 silmultaneous equations.

(b) Looking at the equation you can see that the centre is (5,2). Find the slope of the line beteen that and the given point, invert it and the change the sign, and you now have the slope of the tangent. Use this and the given point to find the equation.

(c) Since it touches both the X and Y axix, g^2=f^2=c.
So, you know have the equation X^2 + Y^2 + 2gx + 2gy + g^2=0.
Put in the given point, and you will find a 2 values for g. Put these values back into the circle equation, and you will have the 2 equations. You should be able to find the radius lengh and centres of each circle from there.

Hope that helps.

Crumbs

Deleted User wrote:

(b) Looking at the equation you can see that the centre is (5,2). Find the slope of the line beteen that and the given point, invert it and the change the sign, and you now have the slope of the tangent. Use this and the given point to find the equation.

That's the method to use if the point (-2,1) is on the circle but it's not. It's outside the circle and we need to find the equation of the two tangents which contain that point.

From the equation, the centre of the circle is (5,2) and the radius is sqrt(10). The general equation of any line containing the point (-2,1) is y - 1 = m(x + 2) which can be rewritten as mx - y + (2m + 1) = 0.

The perpendicular distance from the centre to the tangents is equal to the radius so use the formula for the perpendicular distance from a point to a line and let it equal to sqrt(10). This should give you two values for m which can be substituted back into the above equation to get the equations of the two tangents.

fearcruach

Thanks alot, they're the methods i used for the part (a) and (b) i think that i have the right answer. Part (c) is giving me trouble so i must have another go at it. Btw does anyone know how to get the square of a complex number like: square root 24 - 10i

Crumbs

Crumbs wrote:

That's the method to use if the point (-2,1) is on the circle but it's not. It's outside the circle and we need to find the equation of the two tangents which contain that point.

From the equation, the centre of the circle is (5,2) and the radius is sqrt(10). The general equation of any line containing the point (-2,1) is y - 1 = m(x + 2) which can be rewritten as mx - y + (2m + 1) = 0.

The perpendicular distance from the centre to the tangents is equal to the radius so use the formula for the perpendicular distance from a point to a line and let it equal to sqrt(10). This should give you two values for m which can be substituted back into the above equation to get the equations of the two tangents.

Ah yes. I misread the question.

Crumbs

fearcruach wrote:

Btw does anyone know how to get the square of a complex number like: square root 24 - 10i

Let your answer be (a + bi), so that means that (a + bi)^2 = 24 - 10i

Square out the left side, then equate real with real and imaginary with imaginary. This gives you two equations in a and b (one linear, one quadratic). Solve them to find your answers but remember that a and b have to be real numbers so disregard any complex value that you might get for them.