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Coordinate Geometry - Equations of Circles

Grade 12IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Definition of a circle: The locus of all points in a plane that are at a fixed distance (radius) from a fixed point (center).

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Standard Form: The equation of a circle with center (h, k) and radius r.

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General Form: Expanding the standard form results in x2+y2+Ax+By+C=0x^2 + y^2 + Ax + By + C = 0.

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Completing the Square: The method used to convert the general form back to standard form to identify the center and radius.

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Circle Properties: The tangent to a circle is perpendicular to the radius at the point of contact.

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Intersection: A line and a circle can intersect at two points (secant), one point (tangent), or zero points, determined by the discriminant of the resulting quadratic equation.

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Chord Property: The perpendicular bisector of any chord of a circle passes through the center of the circle.

📐Formulae

Standard Equation: (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

General Equation: x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0

Center (General Form): (−g,−f)(-g, -f)

Radius (General Form): r=g2+f2−cr = \sqrt{g^2 + f^2 - c}

Distance Formula: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Midpoint Formula: M=(x1+x22,y1+y22)M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

💡Examples

Problem 1:

Find the center and radius of the circle given by the equation x2+y2−4x+10y−7=0x^2 + y^2 - 4x + 10y - 7 = 0.

Solution:

Center: (2,−5)(2, -5), Radius: 66

Explanation:

  1. Group x and y terms: (x2−4x)+(y2+10y)=7(x^2 - 4x) + (y^2 + 10y) = 7. 2. Complete the square: (x−2)2−4+(y+5)2−25=7(x - 2)^2 - 4 + (y + 5)^2 - 25 = 7. 3. Simplify: (x−2)2+(y+5)2=7+4+25⇒(x−2)2+(y+5)2=36(x - 2)^2 + (y + 5)^2 = 7 + 4 + 25 \Rightarrow (x - 2)^2 + (y + 5)^2 = 36. 4. Compare with (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2 to find center (2,−5)(2, -5) and r=36=6r = \sqrt{36} = 6.

Problem 2:

Find the equation of the tangent to the circle (x−3)2+(y+1)2=25(x - 3)^2 + (y + 1)^2 = 25 at the point (7,2)(7, 2).

Solution:

4x + 3y - 34 = 0

Explanation:

  1. The center of the circle is C(3,−1)C(3, -1). 2. Find the gradient of the radius (mrm_r) connecting C(3,−1)C(3, -1) and P(7,2)P(7, 2): mr=2−(−1)7−3=34m_r = \frac{2 - (-1)}{7 - 3} = \frac{3}{4}. 3. The tangent is perpendicular to the radius, so its gradient (mtm_t) is −43-\frac{4}{3}. 4. Use the point-slope form: y−2=−43(x−7)y - 2 = -\frac{4}{3}(x - 7). 5. Multiply by 3 and rearrange: 3y−6=−4x+28⇒4x+3y−34=03y - 6 = -4x + 28 \Rightarrow 4x + 3y - 34 = 0.

Problem 3:

Determine if the line y=x+10y = x + 10 intersects the circle x2+y2=25x^2 + y^2 = 25.

Solution:

No intersection.

Explanation:

  1. Substitute y=x+10y = x + 10 into the circle equation: x2+(x+10)2=25x^2 + (x + 10)^2 = 25. 2. Expand: x2+x2+20x+100=25⇒2x2+20x+75=0x^2 + x^2 + 20x + 100 = 25 \Rightarrow 2x^2 + 20x + 75 = 0. 3. Calculate the discriminant (D=b2−4acD = b^2 - 4ac): 202−4(2)(75)=400−600=−20020^2 - 4(2)(75) = 400 - 600 = -200. 4. Since D<0D < 0, there are no real solutions, meaning the line does not intersect the circle.