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C can have a linear term but the constant term is negative. D can have a linear term but the constant term is positive. Hence, it has no linear term and the constant term is negative.

Question 11 Which of the following is not the graph of a quadratic polynomial? These curves are called parabolas. So, option D cannot be possible. Also, the curve of a quadratic polynomial crosses the X-axis at most two point but in option D , the curve crosses the X-axis at the three points, so it does not represent the quadratic polynomial. So, division algorithm is not satisfied.

So the given statement is not correct. Justify your answers. As, a polynomial of degree more than 2 is possible which intersects the x-axis at exactly two points when it has two real roots and other imaginary roots.

Then, product of zeroes Given that, all three zeroes are positive. So, the product of all three zeroes is also positive. Thus, their sum is also positive.

A tangent XY is drawn at point A. We know that, the tangent at any point of a circle is perpendicular to the radius through the point of contact. Join OA and OP. Question 2 The length of tangent from an external point on a circle is always greater than the radius of the circle. Solution: False Because the length of tangent from an external point P on a circle may or may not be greater than the radius of the circle.

Question 3 The length of tangent from an external point P on a circle with centre O is always less than OP. Solution: True PT is a tangent drawn from an external point P. In right angled triangle, hypotenuse is always greater than any of the two sides of the triangle.

Solution: True This may be possible only when both tangent lines coincide or are parallel to each other. Solution: True From point P, two tangents are drawn. Solution: False From point P, two tangents are drawn.

Question 8 If a number of circles touch a given line segment PQ at a point A, Ncert Solutions Of Class 10th Maths Chapter 10 University then their centres lie on the perpendicular bisector of PQ.

To prove: Centres of the circles lie on the perpendicular bisector of PQ. Joining each centre of the circles to the point A on the line segment PQ by line segment i. We know that, if we draw a line from the centre of a circle to its tangent line, then the line is always perpendicular to the tangent line. But it does not bisect the line segment PQ.

Question 9 If a number of circles pass through the end points P and Q of a line segment PQ, then their centres lie on the perpendicular bisector of PQ. We know that the perpendi-cular bisector of a chord of circle always passes through the centre of the circle.

Thus, perpendicular bisector of PQ passes through C 1 and C 2. Similarly, all the circle passing through the end points of line segment PQ, will have their centres on the perpendicular bisector of PQ. Question 1 Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.

Solution: Let C 1 and C 2 be the two circles having same centre O. AC is a chord which touches C 1 at point D. Join OD. Prove that QORP is a cyclic quadrilateral. Since, PR and PQ are tangents. Since, BC and BD are tangents. Question 4 Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of angle formed by tangents.. To prove: Centre of a circle touching two intersecting lines lies on the angle bisector of angle formed by tangents.

Question 8 A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ.

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