﻿ A square is formed by following two pairs of straight lines y2 – 14y + 45 = 0 and x2 – 8x + 12 = 0. A circle is inscribed in it. The centre of the circle is  : Kaysons Education

# A Square Is Formed By Following Two Pairs Of Straight Lines y2 – 14y + 45 = 0 And x2 – 8x + 12 = 0. A Circle Is Inscribed In It. The Centre Of The Circle Is

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## Question

### Solution

Correct option is

(4, 7)

The four lines are x = 2, x = 6, y = 5, y = 9. Hence the vertices of the square are (2, 5), (2, 9), (6, 5), (6, 9). The centre of the circle in the square is mid-point of AC or BD, i.e., (4, 7).

#### SIMILAR QUESTIONS

Q1

A square is inscribed in the circle x2 + y2 – 2x + 4y + 3 = 0. Its sides are parallel to the co-ordinates axes. Then one vertex of the square is

Q2

If the lines 3x – 4y – 7 = 0 and 2x – 3y – 5 = 0 are two diameters of a circle of area 49π square units, then the equation of the circle is:

Q3

A variable circle passes through a fixed point A(pq) and touches the x-axis. The locus of the other end of the diameter through A is:

Q4

A circle touches the x-axis and also touches the circle with centre (0, 3) and radius 2. The locus of the centre of the circle is:

Q5

The triangle PQR is inscribed in the circle x2 + y2 = 25. If Q and R have co-ordinates (3, 4) and (–4, 3) respectively than QPR is equal  to

Q6

Let AB be a chord of the circle x2 + y2 = a2 subtending a right angle at the centre. Then the locus of the centroid of the triangle PAB as Pmoves on the circle is

Q7

The lines joining the origin to the points of intersection of the line 4x + 3y = 24 with the circle (x – 3)2 + (y – 4)2 = 25 are

Q8

If the circles x2 + y2 + 2ax + cy + a = 0 and x2 + y2 – 3ax + dy – 1= 0 intersect in two distinct points P and Q then the line

5x + by – a = 0 passes through P and Q for:

Q9

The intercept on the line y = x by the circle x2 + y2 – 2x = 0 is AB. Equation of the circle on AB as diameter is:

Q10

Let PQ and RS be tangents at the extremities the diameter PR of a circle of radius r. If PS and RQ intersect at a point X on the circumference of the circle, then 2r equals