There is a circle on the coordinate plane with two tangent lines x-y-3 and x-y-0- The points of contacts of the lines with the circle are 3-2- 3-2 and 0-3- Find the radius and the center of the circle-A- 4-5 - -2-3 - 4-9 - 2B- 3 -2-3 - 4-9 - 4C- 9 -2-3 - 2-3 - 4D- 3 - 4 -10-3 - 4-3 - 4

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Basic Properties of a Circle

There is a ci...

Question

There is a circle on the coordinate plane with two tangent lines x - y = 3 and x - y = 0. The points of contacts of the lines with the circle are ( $\frac{3}{2}$ , $\frac{3}{2}$ ) and (0,- 3). Find the radius and the center of the circle.

(3/4)√10,(3/4, -3/4)

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9√2, (3/2, 3/4)

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3√2, (3/4, 9/4)

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4.5/√2 , (3/4, 9/2)

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Solution

The correct option is A
(3/4)√10,(3/4, -3/4)

Let us first look at the nature of the two lines. We observe that the coefficients of x and y for both the lines in equations are same, but the constant term is different for both terms. This means that the lines are parallel.

We know that the line joining the points of contact of parallel tangents is perpendicular to both the tangents and passes through the center of the circle.

$∴$ the centre of the circle is the midpoint of the given points, $(\frac{3}{2}, \frac{3}{2})$ and $(0, - 3)$ , which is;

Centre = $(\frac{\frac{3}{2} + 0}{2}, \frac{\frac{3}{2} - 3}{2})$

= $(\frac{3}{4}, \frac{- 3}{4})$

The radius of the circle would be the distance between the centre, $(\frac{3}{4}, \frac{- 3}{4})$ and a point of contact, $(0, - 3)$ .

Using the distance formula, we can find this distance as:

Radius = $\sqrt{(\frac{3}{4} - 0)^{2} + (\frac{- 3}{4} - 3)^{2}} = (\frac{3}{4}) \sqrt{10}$

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Similar questions

There is a circle on the co -ordinate plane with two tangent lines x - y = 3 and x - y = 0. The meeting points of the circle with the 2 lines are ( $\frac{3}{2}$ , $\frac{3}{2}$ ) and (0,- 3). Find the radius and the centre of the circle.

There is a circle on the co- ordinate plane for which there are 2 tangent lines. These are x - y = 3 and x - y = 0. The meeting points of the circle with the 2 lines are ( $\frac{3}{2}$ , $\frac{3}{2}$ ) and (0,3). Find the radius and the centre of the circle.

Q. Find the equation of the circle which circumscribes the triangle formed by the lines
(i) x + y + 3 = 0, x − y + 1 = 0 and x = 3
(ii) 2x + y − 3 = 0, x + y − 1 = 0 and 3x + 2y − 5 = 0
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Find the equation of the circle which circumscribes the triangle formed by the lines:
(i) $x + y + 3 = 0, x - y + 1 = 0$ and $x = 3$
(ii) $2 x + y - 3 = 0, x + y - 1 = 0$ and $3 x + 2 y - 5 = 0$
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The centre of the circle (x) $^{2}$ +(y) $^{2}$ – 4x – 6y – 12 = 0 is

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There is a circle on the coordinate plane with two tangent lines x - y = 3 and x - y = 0. The points of contacts of the lines with the circle are (32,32) and (0,- 3). Find the radius and the center of the circle.

There is a circle on the coordinate plane with two tangent lines x - y = 3 and x - y = 0. The points of contacts of the lines with the circle are ( $\frac{3}{2}$ , $\frac{3}{2}$ ) and (0,- 3). Find the radius and the center of the circle.