Chapter 12: Q. 5 (page 798)

Find the standard equation of the circle whose center is the point $(- 1, 2)$ if $(3, 5)$ is a point on the circle.

Short Answer

Expert verified

The equation of circle is $x^{2} + y^{2} + 2 x - 4 y - 1 = 0$

Step by step solution

Step 1. Given Information

Centre of circle $(h, k) = (- 1, 2)$ and $(3, 5)$ is on the circle.

Step 2. Finding radius

The given points are $(- 1, 2)$ and $(3, 5)$

$r = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} r = \sqrt{{(3 + 1)}^{2} + {(5 - 2)}^{2}} r = \sqrt{16 + 9} = \sqrt{25} = 5$

Step 3. Finding Equation of Circle

The given centre is $(h, k) = (- 1, 2)$

${(x + 1)}^{2} + {(y - 2)}^{2} = 5 x^{2} + 2 x + 1 + y^{2} - 4 y + 4 = 5 x^{2} + y^{2} + 2 x - 4 y - 1 = 0$

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Find the standard equation of the circle whose center is the point $(- 1, 2)$ if $(3, 5)$ is a point on the circle.

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Finding radius

Step 3. Finding Equation of Circle

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Find the standard equation of the circle whose center is the point -1,2if 3,5is a point on the circle.

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Finding radius

Step 3. Finding Equation of Circle

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Mechanics Maths

Applied Mathematics

Calculus

Theoretical and Mathematical Physics

Geometry

Study anywhere. Anytime. Across all devices.

Find the standard equation of the circle whose center is the point $(- 1, 2)$ if $(3, 5)$ is a point on the circle.