Premium Online Home Tutors

3 Tutor System

Starting just at 265/hour

Answer :

Let the coordinate of centre O be (x,y)

Then, OA = OB = OC {\(\because \) radius is equal}

Now,

OA = \( \sqrt{ (x-6)^2+(y+6)^2 } \)

OB = \( \sqrt{(x-3)^2 + (y+7)^2 } \)

OC = \( \sqrt{(x-3)^2 + (y-3)^2 } \)

Now, OA = OB

\( \Rightarrow \sqrt{(x-6)^2 + (y+6)^2 } = \sqrt{(x-3)^2 + (y+7)^2 } \)

\(\Rightarrow x^2 \ + \ 36 \ - \ 12x \ + \ y^2 \ + \ 36 \ - \ 12 \ = \ x^2 \ + \ 9 \ - \ 6x \ + \ y^2 \ + \ 9 \ - \ 6y \)

\( \Rightarrow 6x \ + 2y \ - \ 14 \ = \ 0 \) ..............(i)

Similarly, OB = OC

\( \Rightarrow \sqrt{(x-3)^2 + (y+7)^2 } = \sqrt{(x-3)^2 + (y-3)^2 } \)

\( \Rightarrow y^2 \ + \ 14y \ + \ 49 \ = \ y^2 \ - \ 6y \ + \ 9 \)

\(\Rightarrow y \ = \ -2 \)

Putting this value of y in eq. (i) , we get

\(\Rightarrow 6x \ + \ 2(-2) \ - \ 14 \ = \ 0 \)

\( \Rightarrow x \ = \ 3 \)

\(\therefore \) The centre of circle is at (3,-2)

- Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, –2) and B(3, 7).
- Find the relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
- The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.
- The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular lawn in the plot as shown in the fig. 7.14. The students are to sow the seeds of flowering plants on the remaining area of the plot. (i) Taking A as origin, find the coordinates of the vertices of the triangle. (ii) What will be the coordinates of the vertices of triangle PQR if C is the origin? Also calculate the areas of the triangles in these cases. What do you observe?
- The vertices of a \( ∆ \ ABC \) are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that \( \frac{AD}{AB} \ = \ \frac{AE}{AC} \ = \ \frac{1}{4} \) . Calculate the area of the \( ∆ \ ADE \) and compare it with area of \( ∆ \ ABC \) . (Recall Theorem 6.2 and Theorem 6.6)
- Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of \( ∆ \ ABC \) . (i) The median from A meets BC at D. Find the coordinates of point D. (ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1. (iii) Find the coordinates of point Q and R on medians BE and CF respectively such that BQ : QE = 2:1 and CR : RF = 2 : 1. (iv) What do you observe? [Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.] (v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, find the coordinates of the centroid of the triangle.
- ABCD is a rectangle formed by the points A (-1, – 1), B (-1, 4), C (5, 4) and D (5, -1). P, Q, R and S are the midpoints of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

- NCERT solutions for class 10 maths chapter 1 Real Numbers
- NCERT solutions for class 10 maths chapter 1 Electricity, Light , Carbon and it's compounds
- NCERT solutions for class 10 maths chapter 2 Polynomials
- NCERT solutions for class 10 maths chapter 3 Pair of linear equations in two variables
- NCERT solutions for class 10 maths chapter 4 Quadratic Equations
- NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions
- NCERT solutions for class 10 maths chapter 6 Triangles
- NCERT solutions for class 10 maths chapter 7 Coordinate Geometry
- NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry
- NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry
- NCERT solutions for class 10 maths chapter 10 Circles
- NCERT solutions for class 10 maths chapter 11 Constructions
- NCERT solutions for class 10 maths chapter 12 Areas related to circles
- NCERT solutions for class 10 maths chapter 13 Surface Areas and Volumes
- NCERT solutions for class 10 maths chapter 14 Statistics
- NCERT solutions for class 10 maths chapter 15 Probability

- NCERT solutions for class 10 science chapter 1 Chemical Reactions and Equations
- NCERT solutions for class 10 science chapter 2 Acids, Bases and Salts
- NCERT solutions for class 10 science chapter 3 Metals and Non Metals
- NCERT solutions for class 10 science chapter 4 Carbon and its Compounds
- NCERT solutions for class 10 science chapter 5 Periodic Classification of Elements
- NCERT solutions for class 10 science chapter 6 Life Processes
- NCERT solutions for class 10 science chapter 7 Control and Coordination
- NCERT solutions for class 10 science chapter 8 How do Organisms Reproduce
- NCERT solutions for class 10 science chapter 9 Heredity and Evolution
- NCERT solutions for class 10 science chapter 10 Light Reflection and Refraction
- NCERT solutions for class 10 science chapter 11 Human Eye and Colorful World
- NCERT solutions for class 10 science chapter 12 Electricity
- NCERT solutions for class 10 science chapter 13 Magnetic Effect of Electric Current
- NCERT solutions for class 10 science chapter 14 Sources of Energy
- NCERT solutions for class 10 science chapter 15 Our Environment
- NCERT solutions for class 10 science chapter 16 Management of Natural Resources