k i am a grade 11 student and need help with some questions

1a) The line y = x - 1 intersects the circle X^2 + y^2 = 25 at two points. This line is called a secant. Find the coordinates of the two points of intersection.

1b) For what value(s) of k will the line y = x + k be a tangent to the circle x^2 + y^2 = 25? A tangent is a line that touches a circle exactly one point.

2) vitaly and jen have 24 m of fencing to enclose a vegetable garden at teh back of their house. What are the dimensions of the largest rectangular garden they could enclose with the length of fencing?
Pic

2. I'll help with the first. So we have the equations:

(I)..$\displaystyle \;y\,=\,x\,-\,1$

(II)..$\displaystyle \;x^2\,+\,y^2\,=\,25$

To find the point of intersection with algebra, we just solve the system for x and y.

So let's substitute the value of y in (I) into (II):

$\displaystyle x^2\,+\,(x\,-\,1)^2\,=\,25$

Expand $\displaystyle (x\,-\,1)^2$:..$\displaystyle x^2\,+\,x^2\,-\,2x\,+\,1\,=\,25$

Simplify:..$\displaystyle 2x^2\,-\,2x\,-\,12=\,0\;\Rightarrow\;x^2\,-\,x\,-\,12\,=\,0$

Factor:..$\displaystyle (x\,-\,4)(x\,+\,3)\,=\,0\;\Rightarrow\;x\,=\,4\,,-\,3$

Now substitute our value of x into (I):

For the x value of 4:..$\displaystyle y\,=\,(4)\,-\,1\,=\,3$

So we have the point:..$\displaystyle (4\,,\,3)$

For the x value of -3, substituting it into (I), we get -4.

So we also have the other point $\displaystyle (-\,3\,,\,-\,4)$

The graph agrees with my answer:

3. Wow Thanks alot for your help
That was really fast

Thanks again

4. 1b)
The system $\displaystyle \displaystyle \left\{\begin{array}{ll}x^2+y^2=25\\y=x+k\end{arra y}\right.$ must have a single solution.
Replacing $\displaystyle y$ in the first equation we obtain $\displaystyle 2x^2+2kx+k^2-25=0$.
This equation must have equal roots, so $\displaystyle \triangle =b^2-4ac=0\Leftrightarrow k^2=50\Rightarrow k=\pm 5\sqrt{2}$

1c)
Let $\displaystyle x,y$ be the length sides of the rectangle.
So we have $\displaystyle 2x+2y=24\Rightarrow x+y=12$. (1)
The area of the rectangle is $\displaystyle A=xy$.
Replacing $\displaystyle y$ from (1) in the area we obtain $\displaystyle A=x(12-x)$.
We can consider the area as a function of variable $\displaystyle x$ and we must find the maximum of this function.
The maximum of a quadratic function $\displaystyle f(x)=ax^2+bx+c,a<0$ is $\displaystyle \displaystyle -\frac{\triangle}{4a}$, where $\displaystyle \triangle =b^2-4ac$.
In this case $\displaystyle f(x)=-x^2+12x$ and the maximum is 36.
This maximum is obtained when $\displaystyle x=y=6$, so the rectangle is a square.

5. Thanks alot for the sloution

I love this forum