NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 10: Construction - Exercise 10.3Question 4
Class 10 - Mathematics - CHAPTER 10: Construction - Exercise 10.3
Question. 4
Construct a tangent to a circle of radius \(4\,\text{cm}\) from a point which is at a distance of \(6\,\text{cm}\) from its centre.
Answer:
- Draw the circle with centre \(O\) and radius \(4\,\text{cm}\).
- Mark a point \(P\) such that \(OP=6\,\text{cm}\).
- Construct the midpoint \(M\) of \(OP\). With centre \(M\) and radius \(MO\), draw a circle; it meets the given circle at \(T\).
- Join \(PT\). Then \(PT\) is the required tangent. (Similarly, the other intersection gives the second tangent.)
Length (by right triangle \(\triangle OPT\)): \[ PT=\sqrt{OP^2-OT^2}=\sqrt{6^2-4^2}=\sqrt{20}=2\sqrt{5}\,\text{cm}. \]
Detailed Answer with Explanation:
Step 1: We know a tangent touches the circle at exactly one point and makes a right angle (90°) with the radius at the point of contact.
Step 2: When we draw a circle with diameter \(OP\), by the property of a circle, the angle made at the circumference (here at point \(T\)) is a right angle, i.e., \(\angle OTP = 90^{\circ}\).
Step 3: Since \(\angle OTP = 90^{\circ}\), line \(PT\) is perpendicular to radius \(OT\). This means \(PT\) only touches the circle at \(T\) and does not cut through it.
Step 4: Therefore, \(PT\) is a tangent to the circle from the external point \(P\).
In simple words: We used the property of a circle drawn with diameter \(OP\) to locate the right angle at \(T\). This gave us the exact point where the tangent from \(P\) touches the circle.