A tangent to acircleis a line that touches the circle at one point.
Tangent Theorem:
1) The tangent to the circle is perpendicular to the radius at the starting point
2) Every line perpendicular to the radius at its end is tangent to the circle
3) The angle between the tangent and any chord is equal to the circumferential angle that rests on that chord on the other side.
4) Two tangents to the circle that come out from the same point are equal to each other.
5) A segment that passes between the center of the circle and the point from which two tangents to the circle come out, cuts the angle between the tangents.
6) If from any point outside the circle, a tangent comes out and cuts the circle, then the product of the entire tangent on its outside is equal to the square of the tangent.
7) In the triangle that encloses the circle, the three bisectors of the angles of the triangle meet at a point in the center of the circle.
8) We can determine that a convex quadrilateral encloses a circle only if - the sum of two opposite sides in the square will be equal to the sum of the other two sides in the square.
In which of the circles is the point marked in the circle and not on the circumference?
Video Solution
Step-by-Step Solution
Let's remember that the circular line draws the shape of the circle, and the inner part is called a disk.
Therefore, in diagram B, the point is located in the inner part, meaning inside the disk.
Answer
Exercise #2
A circle has the following equation: x2−8ax+y2+10ay=−5a2
Point O is its center and is in the second quadrant (a=0)
Use the completing the square method to find the center of the circle and its radius in terms of a.
Step-by-Step Solution
Let's recall that the equation of a circle with its center at O(xo,yo) and its radius R is:
(x−xo)2+(y−yo)2=R2Now, let's now have a look at the equation for the given circle:
x2−8ax+y2+10ay=−5a2 We will try rearrange this equation to match the circle equation, or in other words we will ensure that on the left side is the sum of two squared binomial expressions, one for x and one for y.
We will do this using the "completing the square" method:
Let's recall the short formula for squaring a binomial:
(c±d)2=c2±2cd+d2We'll deal separatelywith the part of the equation related to x in the equation (underlined):
x2−8ax+y2+10ay=−5a2
We'll isolate these two terms from the equation and deal with them separately.
We'll present these terms in a form similar to the form of the first two terms in the shortcut formula (we'll choose the subtraction form of the binomial squared formula since the term in the first power we are dealing with is8ax, which has a negative sign):
x2−8ax↔c2−2cd+d2↓x2−2↓⋅x⋅4a↔c2−2↓cd+d2Notice that compared to the short formula (which is on the right side of the blue arrow in the previous calculation), we are actually making the comparison:
{x↔c4a↔d Therefore, if we want to get a squared binomial form from these two terms (underlined in the calculation), we will need to add the term(4</span><spanclass="katex">a)2, but we don't want to change the value of the expression, and therefore we will also subtract this term from the expression.
That is, we will add and subtract the term (or expression) we need to "complete" to the binomial squared form,
In the following calculation, the "trick" is highlighted (two lines under the term we added and subtracted from the expression),
Next, we'll put the expression in the squared binomial form the appropriate expression (highlighted with colors) and in the last stage we'll simplify the expression:
x2−2⋅x⋅4ax2−2⋅x⋅4a+(4a)2−(4a)2x2−2⋅x⋅4a+(4a)2−16a2↓(x−4a)2−16a2Let's summarize the steps we've taken so far for the expression with x.
We'll do this within the given equation:
x2−8ax+y2+10ay=−5a2x2−2⋅x⋅4a+(4a)2−(4a)2+y2+10ay=−5a2↓(x−4a)2−16a2+y2+10ay=−5a2We'll continue and do the same thing for the expressions with y in the resulting equation:
(Now we'll choose the addition form of the squared binomial formula since the term in the first power we are dealing with 10ay has a positive sign)
(x−4a)2−16a2+y2+10ay=−5a2↓(x−4a)2−16a2+y2+2⋅y⋅5a=−5a2(x−4a)2−16a2+y2+2⋅y⋅5a+(5a)2−(5a)2=−5a2↓(x−4a)2−16a2+y2+2⋅y⋅5a+(5a)2−25a2=−5a2↓(x−4a)2−16a2+(y+5a)2−25a2=−5a2(x−4a)2+(y+5a)2=36a2In the last step, we move the free numbers to the second side and combine like terms.
Now that the given circle equation is in the form of the general circle equation mentioned earlier, we can easily extract both the center of the given circle and its radius:
In the last step, we made sure to get the exact form of the general circle equation—that is, where only subtraction is performed within the squared expressions (emphasized with an arrow)
Therefore, we can conclude that the center of the circle is at:O(xo,yo)↔O(4a,−5a) and extract the radius of the circle by solving a simple equation:
R2=36a2/→R=±6a
Remember that the radius of the circle, by its definition is the distance between any point on the diameter and the center of the circle. Since it is positive, we must disqualify one of the options we got for the radius.
To do this, we will use the remaining information we haven't used yet—which is that the center of the given circle O is in the second quadrant.
That is:
O(x_o,y_o)\leftrightarrow x_o<0,\hspace{4pt}y_o>0 (Or in words: the x-value of the circle's center is negative and the y-value of the circle's center is positive)