Class 10: Circles – Exercise 8.1

$\displaystyle \textbf{Question 1: }\text{Fill in the blanks:}$
$\displaystyle \text{(i) The common point of a tangent and the circle is called }\underline{\hspace{1.5cm}}\text{.}$
$\displaystyle \text{(ii) A circle may have }\underline{\hspace{1.5cm}}\text{ parallel tangents.}$
$\displaystyle \text{(iii) A tangent to a circle intersects it in }\underline{\hspace{1.5cm}}\text{ point(s).}$
$\displaystyle \text{(iv) A line intersecting a circle in two points is called a }\underline{\hspace{1.5cm}}\text{.}$
$\displaystyle \text{(v) The angle between tangent at a point on a circle and the radius through the point is }\underline{\hspace{1.5cm}}\text{.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Point of contact}$
$\displaystyle \text{The common point of a tangent and the circle is called the point of contact.}$
$\displaystyle \therefore \text{The required answer is point of contact.}$

$\displaystyle \text{(ii) Two}$
$\displaystyle \text{For a given direction, a circle can have two parallel tangents, one on each side.}$
$\displaystyle \therefore \text{The required answer is two.}$

$\displaystyle \text{(iii) One}$
$\displaystyle \text{By definition, a tangent intersects a circle at exactly one point.}$
$\displaystyle \therefore \text{The required answer is one.}$

$\displaystyle \text{(iv) Secant}$
$\displaystyle \text{A line which intersects a circle at two distinct points is called a secant.}$
$\displaystyle \therefore \text{The required answer is secant.}$

$\displaystyle \text{(v) }90^\circ$
$\displaystyle \text{The tangent at a point on a circle is perpendicular to the radius through that point.}$
$\displaystyle \therefore \text{The angle between the tangent and the radius}=90^\circ.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{How many tangents can a circle have?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{A tangent can be drawn at every point on a circle.}$
$\displaystyle \text{Since a circle has infinitely many points, it can have infinitely many tangents.}$
$\displaystyle \therefore \text{A circle can have infinitely many tangents.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{O is the centre of a circle of radius }8\text{ cm. The tangent at a point }A$
$\displaystyle \text{on the circle cuts a line through }O\text{ at }B\text{ such that }AB=15\text{ cm. Find }OB\text{.}$
$\displaystyle \text{Answer:}$
$\displaystyle OA=8\text{ cm},\qquad AB=15\text{ cm}$
$\displaystyle \text{Since the radius through the point of contact is perpendicular to the tangent,}$
$\displaystyle OA\perp AB.$
$\displaystyle \text{Therefore, }\triangle OAB\text{ is right-angled at }A.$
$\displaystyle OB^2=OA^2+AB^2$
$\displaystyle =8^2+15^2=64+225=289$
$\displaystyle \therefore OB=\sqrt{289}=17\text{ cm}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{If the tangent at a point }P\text{ to a circle with centre }O\text{ cuts a line through }O$
$\displaystyle \text{at }Q\text{ such that }PQ=24\text{ cm and }OQ=25\text{ cm, find the radius of the circle.}$
$\displaystyle \text{Answer:}$
$\displaystyle PQ=24\text{ cm},\qquad OQ=25\text{ cm}$
$\displaystyle \text{Since the radius through the point of contact is perpendicular to the tangent,}$
$\displaystyle OP\perp PQ.$
$\displaystyle \text{Therefore, }\triangle OPQ\text{ is right-angled at }P.$
$\displaystyle OQ^2=OP^2+PQ^2$
$\displaystyle 25^2=OP^2+24^2$
$\displaystyle OP^2=625-576=49$
$\displaystyle \therefore OP=\sqrt{49}=7\text{ cm}.$
$\displaystyle \therefore \text{The radius of the circle is }7\text{ cm}.$
$\displaystyle \\$